Limits and Derivatives | ISC Class 11 Maths Notes

This note covers limits, left-hand and right-hand limits, algebraic and trigonometric limits, exponential and logarithmic limits, indeterminate forms, derivatives as rates of change, tangent slopes, differentiation from first principles, and rules for differentiating sums, differences, products and quotients.

What does the limit of a function mean?

Calculus studies change in the value of a function as its input changes. A function assigns one output to each allowed input. Its domain is the set of allowed inputs. Write f(x) for the output of function f at input x.

Let a be a fixed real number, meaning a number on the number line, and let L denote a proposed limiting value. The notation x → a means that x approaches a. The symbol → is read as “tends to”.

Definition: A limit describes the value approached by f(x) as x approaches a through nearby points of its domain. We write $\lim_{x\to a}f(x)=L$, read as “the limit of f(x), as x tends to a, is L”.

How is a limit different from a function value?

The function value f(a) uses the input a itself. The limit concerns inputs near a. These ideas must be kept separate: either may exist without the other, and they need not be equal when both exist.

For f(x) = x², where x² means x multiplied by itself, the limit as x → 0 is 0. Here f(0) also equals 0. However, agreement in this example does not make equality with the function value part of the definition of a limit.

Worked example 1. Find the limit as x → 2 of f(x) = (x² − 4)/(x − 2), defined for x ≠ 2. The symbol ≠ means “is not equal to”.

Answer: Factor x² − 4 as (x − 2)(x + 2). At allowed inputs, cancelling the non-zero factor x − 2 gives f(x) = x + 2. Hence the limit is 4, although f(2) is undefined.

The graph has a straight rising line with a hollow point at (2, 4). Dashed guides connect that point to (2, 0) and (0, 4) on the axes. It also labels the intercepts, where the line crosses the axes, at (−2, 0) and (0, 2).

See Fig. 12.2 in your NCERT textbook

In a coordinate pair (x, y), x is the horizontal coordinate and y is the vertical coordinate, representing the function output. The hollow point shows why approaching an output of 4 does not assign a value to the function at the missing input.

When do left-hand and right-hand limits give a limit?

At a point with domain values on both sides, x may approach a through values less than a or through values greater than a. These approaches define the left-hand limit and right-hand limit, respectively. Examine both when the function has different rules on different sides.

Write $\lim_{x\to a^-}f(x)$ for the left-hand limit and $\lim_{x\to a^+}f(x)$ for the right-hand limit. The superscript minus and plus indicate the side of approach. They do not mean subtracting or adding a fixed number to a.

Result: The two-sided limit criterion

The two-sided limit exists as a finite real number if the left-hand and right-hand limits exist and are equal. The limit is their common value. If the two sides give different values, there is no two-sided limit at that point.

Worked example 2. Let f(x) = 1 for x ≤ 0 and f(x) = 2 for x > 0. Find the left-hand limit, right-hand limit and two-sided limit at 0. Here ≤ means “less than or equal to”, and > means “greater than”.

Answer: For negative inputs near 0, f(x) remains 1, so the left-hand limit is 1. For positive inputs near 0, f(x) remains 2, so the right-hand limit is 2. They differ, so the two-sided limit does not exist, although f(0) = 1.

Two horizontal pieces lie at heights 1 and 2. The lower piece lies to the left of the vertical axis and ends at the filled point (0, 1). The upper piece extends rightwards from the hollow point (0, 2).

See Fig. 12.3 in your NCERT textbook

Can both limits exist but differ from the function value?

Yes. Consider f(x) = x + 2 for x ≠ 1, with f(1) = 0. Inputs near 1 follow the rule x + 2 from either side. Both limits equal 3, while the assigned function value is 0.

A piecewise function uses different formulae for specified parts of its domain. Read the conditions before substituting. The formula assigned exactly at the joining point does not determine what happens on either side of it.

How do the fundamental laws simplify limits?

The algebra of limits allows a complicated expression to be separated into simpler functions. Its hypotheses matter: the limits being combined must exist. For a quotient, the limit of the denominator must also be non-zero. Checking the denominator is part of using the rule.

Theorem: Sum, difference, product and quotient laws

Let f and g be functions with finite limits A and B, respectively, as x → a. Here g is a second function, A is the limit of f, and B is the limit of g: $\lim_{x\to a}f(x)=A$ and $\lim_{x\to a}g(x)=B$.

Juxtaposition, as in AB, denotes multiplication. If c is a fixed real constant, the constant-multiple rule gives $\lim_{x\to a}[cf(x)]=cA$. A constant has a value that does not change when x changes.

How do these laws apply to polynomials?

A polynomial is a finite sum of constant multiples of non-negative whole-number powers of x. Its degree is the largest power with a non-zero coefficient, where a coefficient is the number multiplying a power. Polynomial limits at finite inputs equal their values there.

Worked example 3. Evaluate $\lim_{x\to1}(x^3-x^2+1)$, where x³ means x multiplied by itself three times.

Answer: The expression is a polynomial, so substitute x = 1 into each term. The limit is 1³ − 1² + 1 = 1. No cancellation is needed because each term has a finite limit.

A rational function is a quotient of polynomials on inputs where its denominator is non-zero. Direct substitution gives its limit when the denominator stays non-zero at the limiting input. If substitution makes that denominator zero, the quotient law cannot yet be used.

Worked example 4. Evaluate $\lim_{x\to1}(x^2+1)/(x+100)$.

Answer: The numerator tends to 2 and the denominator tends to 101. Since 101 ≠ 0, the quotient law applies and gives the limit 2/101.

How are indeterminate algebraic limits evaluated?

An indeterminate form is a form that does not by itself determine the limit. The form 0/0 occurs when direct substitution makes both numerator and denominator zero. It is neither the number zero nor a licence to divide by zero.

Look for an equivalent expression valid near the limiting input. Factorisation rewrites an expression as a product. Cancelling a common non-zero factor may reveal a quotient to which the limit laws apply. Cancellation concerns nearby inputs, so it does not fill a hole in the original domain.

Worked example 5. Evaluate $\lim_{x\to2}(x^3-2x^2)/(x^2-5x+6)$.

Answer: Direct substitution gives 0/0. Factor the numerator as x²(x − 2) and the denominator as (x − 2)(x − 3). For x ≠ 2, cancel x − 2. The remaining expression is x²/(x − 3), whose limit is 4/(−1) = −4.

Result: A standard difference-of-powers limit

Let n be a positive integer, meaning a whole number greater than zero. Then $\lim_{x\to a}(x^n-a^n)/(x-a)=na^{n-1}$. The formula also holds for rational n when a is positive; a rational number is a ratio of integers with non-zero denominator.

The positive-integer result follows by cancelling x − a in the difference of powers. What remains is a sum of n terms; each approaches the same power of a. The result is useful when large powers make full expansion inconvenient.

Worked example 6. Evaluate $\lim_{x\to1}(x^{15}-1)/(x^{10}-1)$.

Answer: Divide both numerator and denominator by x − 1 for nearby x ≠ 1. The expression becomes [(x¹⁵ − 1)/(x − 1)] divided by [(x¹⁰ − 1)/(x − 1)]. Their limits are 15 and 10, so the required limit is 15/10 = 3/2.

How can a square root be handled?

The square root symbol √ denotes the non-negative number whose square equals the expression inside the root. For $\lim_{x\to0}(\sqrt{1+x}-1)/x$, put y = 1 + x, where y is a new input variable. Then y → 1, and the rational-power result with n = 1/2 gives the limit 1/2.

Note: Getting 0/0 is a signal to simplify, not a final answer. A denominator becoming zero does not automatically rule out a finite limit: cancellation may remove the obstruction, but the remaining expression must still be examined.

Which standard limits are used for trigonometric functions?

Trigonometric functions assign numbers to angles. Their angle inputs here are measured in radians. One radian is the angle at a circle's centre subtended by an arc whose length equals the radius.

On a circle of radius 1 centred at the origin (0, 0), measure angle x from the positive horizontal axis. The horizontal and vertical coordinates of its point on the circle are cosine, written cos x, and sine, written sin x, respectively.

Tangent, written tan x, is sin x/cos x where cos x ≠ 0. Positive angles are measured anticlockwise; negative angles are measured clockwise. These trigonometric values therefore apply to angles of either sign.

The number π is the ratio of a circle's circumference to its diameter; a straight angle is π radians. A degree is 1/360 of a full turn. Radian measure is essential in the standard limits below. Changing to degree measure changes the numerical relationship between the angle and its sine.

Theorem: Standard trigonometric limits

As x approaches zero through non-zero values, $\lim_{x\to0}\sin x/x=1$ and $\lim_{x\to0}(1-\cos x)/x=0$. Also, since tan x = sin x/cos x and cos x → 1, $\lim_{x\to0}\tan x/x=1$.

The sandwich theorem states that a function bounded between two functions with the same limit has that limit. For positive x smaller than π/2, cos x < sin x/x < 1, where < means “less than”. Both bounds approach 1 as x → 0.

The corresponding conclusion from negative inputs follows because sin(−x) = −sin x and cos(−x) = cos x. Thus the one-sided conclusions agree. The important condition is agreement of the bounding limits, not merely the presence of an inequality.

Worked example 7. Evaluate $\lim_{x\to0}\sin(4x)/\sin(2x)$, with angles in radians.

Answer: Rewrite the expression as 2[sin(4x)/(4x)]/[sin(2x)/(2x)]. As x → 0, both 4x and 2x approach zero. Each bracketed quotient tends to 1, so the required limit is 2 × 1/1 = 2.

Why must the denominator match the angle?

The standard sine limit pairs sin x with the same x in the denominator. For sin(4x), create the denominator 4x before applying it. Any factors inserted during this rearrangement must be balanced outside the quotient.

An identity is an equality valid for every permitted input. The identity 1 − cos x = 2 sin²(x/2), where sin² means the square of sine, converts cosine expressions into sine limits. It gives (1 − cos x)/x = [sin(x/2)/(x/2)] sin(x/2), whose factors tend to 1 and 0.

How are exponential and logarithmic limits recognised?

An exponential function has the variable in its exponent, the power to which a base is raised. In bˣ, b is the fixed positive base and x is the exponent. For logarithms take b ≠ 1. The logarithm log to base b of a positive number is the exponent that produces that number from b.

Thus $\log_b x=y$ means $b^y=x$, with x > 0. The constant e is the base of the natural exponential function eˣ. It is the number given by the series 1 + 1/1! + 1/2! + …, lying between 2 and 3.

Here the dots mean the series continues, and the factorial notation n! means the product of the positive integers from 1 to n. The natural logarithm, written ln x, is the logarithm to base e. Keep ln distinct from a logarithm with another base.

Result: Exponential and logarithmic limit forms

These are standard limits: direct substitution gives 0/0, so ordinary division of the substituted values does not evaluate them. The base affects the answer. In particular, the value 1 in the logarithmic formula above belongs to the natural logarithm.

How are ordinary limits different from these quotient forms?

For a fixed real a, $\lim_{x\to a}b^x=b^a$ when b > 0. For a > 0, $\lim_{x\to a}\ln x=\ln a$. These substitution rules apply where the functions are defined, unlike the quotient forms requiring a limiting calculation.

The change-of-base rule $\log_b x=\ln x/\ln b$ explains the logarithmic factor 1/ln b. Apply it before evaluating the limit, and keep the condition b ≠ 1: otherwise ln b is zero and the formula divides by zero.

Note: Check the logarithm's argument, the expression inside it. For ln(1 + x), that argument must be positive. Near zero this allows approach from either side, provided x remains greater than −1.

What does a derivative measure geometrically and physically?

A derivative measures the instantaneous rate of change of a function with respect to its input. “Instantaneous” refers to change at a particular input, obtained as a limit of average rates over shrinking intervals. A finite average rate and a derivative are different quantities.

Let h be a non-zero change in input. Then f(a + h) − f(a) is the corresponding change in output. Their ratio, [f(a + h) − f(a)]/h, is the difference quotient, representing the average rate over that change.

Definition: The derivative of f at a is $f'(a)=\lim_{h\to0}[f(a+h)-f(a)]/h$, provided this finite limit exists. The prime symbol ′ denotes a derivative. The function must be defined at a and at the nearby inputs used.

How does the difference quotient become a tangent slope?

On the graph y = f(x), label P = (a, f(a)) and Q = (a + h, f(a + h)). The secant through P and Q joins two points of the curve. Its slope is vertical change divided by horizontal change, namely the difference quotient.

As h → 0, Q approaches P. When the limiting slope exists, it gives the slope of the tangent, the limiting position of the secant at P. Therefore f′(a) is the slope of that tangent.

A curve labelled y = f(x) contains P(a, f(a)) and Q(a + h, f(a + h)). Dashed horizontal and vertical guides form the right-angled triangle PQR. R is the point directly below Q at P's height; PR is labelled h.

See Fig. 12.11 in your NCERT textbook

How does this describe a moving body's velocity?

Let s denote distance in metres and t denote elapsed time in seconds. For the distance function s = 4.9t², average velocities over progressively smaller intervals around t = 2 suggest the body's velocity at that instant.

The numerical estimates from either side place that velocity between 19.551 and 19.649 metres per second. This is an estimate from nearby values, not an exact calculation of the derivative. The limiting approach connects the physical rate of change with the tangent slope.

Once differentiation has been performed, a derivative at a particular input is a number. The derivative function gives the derivative at each input where the defining limit exists. These uses are related, but the input has been fixed in the first and remains variable in the second.

How is differentiation carried out from first principles?

Differentiation is the process of finding a derivative. From first principles, use the difference quotient and evaluate its limit. For a variable input x, the definition is $f'(x)=\lim_{h\to0}[f(x+h)-f(x)]/h$, wherever the finite limit exists.

The alternative notation df/dx means the derivative of f with respect to x. If y = f(x), dy/dx denotes the same derivative. The symbol d/dx is the instruction to differentiate the expression that follows with respect to x.

What order should the calculation follow?

1. Write the given function f(x) and calculate f(x + h) by replacing each occurrence of x with x + h.
2. Subtract the whole expression f(x), using brackets so every sign is handled correctly.
3. Divide the difference by h, and simplify while h ≠ 0.
4. Take the limit as h → 0, then substitute a specified input if a derivative at a point is required.

Worked example 8. Differentiate f(x) = x² from first principles.

Answer: f(x + h) − f(x) = (x + h)² − x² = 2xh + h². Dividing by h gives 2x + h for h ≠ 0. Taking h → 0 gives f′(x) = 2x.

Cancellation before taking the limit is the crucial step. Substituting h = 0 into the unsimplified difference quotient gives 0/0. The calculation instead considers non-zero increments approaching zero, just as cancellation in an algebraic limit uses inputs near the excluded point.

Worked example 9. Differentiate f(x) = 1/x from first principles, for x ≠ 0.

Answer: f(x + h) − f(x) = 1/(x + h) − 1/x = −h/[x(x + h)]. Division by h gives −1/[x(x + h)]. Taking h → 0 gives f′(x) = −1/x², with the original restriction x ≠ 0 retained.

What happens for a constant function?

If f(x) = c, then f(x + h) − f(x) = c − c = 0. The difference quotient is zero for every non-zero h, so f′(x) = 0. A constant output has no change as the input varies.

For f(x) = 3x at x = 2, the quotient is [3(2 + h) − 3(2)]/h = 3h/h = 3. Its limit is 3. Thus the derivative can be evaluated directly at a specified input without first finding a general formula.

How do derivative rules simplify polynomial and rational functions?

Let u and v denote differentiable functions of x, meaning that their derivatives exist at the inputs under consideration. Write u′ and v′ for their derivatives. The rules below combine these functions without repeating the full first-principles calculation each time.

Theorem: Algebra of derivatives

The product rule differentiates one factor at a time and adds the resulting terms. It is not the product of the two derivatives. The quotient rule keeps the order u′v − uv′ in its numerator and squares the original denominator.

Result: The power and constant-multiple rules

For a positive integer n, $d(x^n)/dx=nx^{n-1}$. This power rule extends to real powers where the function and derivative are defined. Also, $d[cu]/dx=cu'$, since the derivative of the constant c is zero.

Differentiate a polynomial term by term using the sum and difference rules. Multiply each coefficient by its power, reduce that power by one, and remove the constant term because its derivative is zero. A linear term contributes its coefficient.

Worked example 10. Differentiate f(x) = 6x¹⁰⁰ − x⁵⁵ + x.

Answer: The derivatives of the three terms are 600x⁹⁹, −55x⁵⁴ and 1. Combining them in their original order gives f′(x) = 600x⁹⁹ − 55x⁵⁴ + 1.

Worked example 11. Differentiate f(x) = (x + 1)/x, where x ≠ 0.

Answer: Put u = x + 1 and v = x. Then u′ = 1 and v′ = 1. The quotient rule gives f′(x) = [x − (x + 1)]/x² = −1/x². The derivative is defined for x ≠ 0.

For a derivative at a specified point, differentiate first and evaluate the resulting formula there. Substituting the point into the original function first turns the output into a constant and loses the information about how nearby outputs change.

How are trigonometric functions differentiated?

With angles in radians, the derivatives of sine and cosine are cos x and −sin x, respectively. The negative sign in the cosine derivative matters. Other trigonometric derivatives follow using these results with the product and quotient rules.

Define sec x, or secant, as 1/cos x; cosec x, or cosecant, as 1/sin x; and cot x, or cotangent, as cos x/sin x. Here “secant” names a trigonometric function, distinct from the secant line joining two curve points.

Result: Standard trigonometric derivatives

The square in sec² x or cosec² x applies to the function value, just as in sin² x. Domain restrictions survive differentiation: a derivative formula does not make the original function defined where its denominator is zero.

How does the sine derivative follow from a limit?

For f(x) = sin x, the difference quotient is [sin(x + h) − sin x]/h. The sine-difference identity rewrites it as cos(x + h/2)[sin(h/2)/(h/2)]. As h → 0, the factors approach cos x and 1.

Worked example 12. Differentiate f(x) = sin² x with x in radians.

Answer: Write sin² x = sin x sin x and use the product rule. Then f′(x) = cos x sin x + sin x cos x = 2 sin x cos x = sin(2x).

Worked example 13. Differentiate f(x) = tan x, where cos x ≠ 0 and x is in radians.

Answer: Write tan x = sin x/cos x. The quotient rule gives [cos² x + sin² x]/cos² x. Since sin² x + cos² x = 1, the derivative is 1/cos² x = sec² x.

For cot x, start from cos x/sin x. The numerator in the quotient rule is −sin² x − cos² x = −1, so its derivative is −1/sin² x = −cosec² x. Writing the unsimplified numerator helps preserve the correct signs.

Glossary

• Function — A rule assigning exactly one output to each allowed input in its domain.
• Domain — The set of input values for which a function is defined.
• Limit — The value approached by a function as its input approaches a specified point.
• Left-hand limit — The limiting value determined by inputs approaching a point from smaller values.
• Right-hand limit — The limiting value determined by inputs approaching a point from greater values.
• Indeterminate form — A form such as zero divided by zero that does not itself determine a limit.
• Rational function — A quotient of polynomials defined at inputs where the denominator is non-zero.
• Radian — An angle at a circle's centre subtended by an arc equal in length to its radius.
• Natural logarithm — The logarithm to base e, written ln, and defined for positive real arguments.
• Difference quotient — The change in a function's output divided by a corresponding non-zero input change.
• Derivative — The finite limit of a difference quotient as the input change approaches zero.
• Tangent slope — The limiting slope of secants through a fixed curve point as another curve point approaches it.
• First principles — The method of differentiating directly by simplifying the difference quotient and evaluating its limit.
• Product rule — A differentiation rule that changes one factor at a time and adds the resulting terms.

Common errors and misconceptions

• Misconception: A limit must equal the function value. Correct: Nearby values determine the limit. The value at the point may differ or may be undefined.
• Misconception: A right-hand limit alone proves a two-sided limit exists. Correct: When the domain extends to both sides, calculate both one-sided limits and check that they agree.
• Misconception: The form 0/0 has value zero. Correct: It is indeterminate. Simplify the expression near the point before deciding whether a finite limit exists.
• Misconception: Cancelling x − a assigns a value at x = a. Correct: Cancellation is valid for x ≠ a. It may establish a limit without defining the original function at a.
• Misconception: The standard sine limit uses any angle unit. Correct: The limit of sin x/x is 1 as x → 0 when x is measured in radians.
• Misconception: Set h = 0 before simplifying a first-principles quotient. Correct: Keep h non-zero during division and cancellation, then take the limit as h approaches zero.
• Misconception: The derivative of uv is u′v′. Correct: Use u′v + uv′. For a quotient, use (u′v − uv′)/v², with v ≠ 0.
• Misconception: A logarithmic limit is independent of its base. Correct: The limit of log to base b of (1 + x), divided by x, is 1/ln b as x → 0.

Exam-style questions with model answers

Q1. Let f(x) = 1 for x ≤ 0 and f(x) = 2 for x > 0. State the one-sided limits at 0 and decide whether the two-sided limit exists. [2 marks]

1. The left-hand limit is 1, while the right-hand limit is 2, because the function is constant on each respective side.
2. These values differ, so the two-sided limit at 0 does not exist, although the given function has f(0) = 1.

Q2. Evaluate the limit of (x³ − 2x²)/(x² − 5x + 6) as x → 2, explaining why cancellation is allowed. [3 marks]

1. Substituting x = 2 makes both numerator and denominator zero, giving the indeterminate form 0/0. This is not a numerical answer.
2. Factor to obtain x²(x − 2)/[(x − 2)(x − 3)]. For nearby x ≠ 2, the common non-zero factor x − 2 can be cancelled.
3. The remaining quotient is x²/(x − 3). Its denominator approaches −1, so direct substitution now gives the limit 4/(−1) = −4.

Q3. Evaluate the limit of sin(4x)/sin(2x) as x → 0, with x measured in radians. Use the standard limit sin u/u → 1 as u → 0, where u denotes an angle. [3 marks]

1. Match each sine with its own angle in the denominator: sin(4x)/sin(2x) = 2[sin(4x)/(4x)]/[sin(2x)/(2x)] for sufficiently small non-zero x.
2. As x approaches zero, both angle inputs 4x and 2x also approach zero. Therefore each bracketed sine quotient tends to 1 by the supplied standard limit.
3. The denominator's limiting value is non-zero, so the quotient law applies. The required limit equals 2 × 1/1 = 2.

Q4. For f(x) = x², derive f′(x) from first principles and explain its meaning as a tangent slope. Here h denotes a non-zero increment in x. [5 marks]

1. Begin with the definition f′(x) = lim as h → 0 of [f(x + h) − f(x)]/h. This measures the limiting change in output per change in input.
2. Substitute the given function to obtain [(x + h)² − x²]/h. The expression uses nearby inputs x + h, with h kept non-zero during the calculation.
3. Expand the square: (x + h)² − x² = 2xh + h². Thus the whole difference quotient becomes (2xh + h²)/h.
4. Cancel the factor h to get 2x + h. Its limit as h → 0 is 2x, giving f′(x) = 2x.
5. The original quotient is the slope of the secant through the two curve points. Its limit is the tangent slope, so the tangent at input x has slope 2x.

Q5. For f(x) = 2x² + 3x − 5, calculate f′(−1) and f′(0), then verify that f′(0) + 3f′(−1) = 0. [3 marks]

1. Differentiate the given polynomial term by term: the derivatives of 2x², 3x and −5 are 4x, 3 and 0. Therefore f′(x) = 4x + 3.
2. Evaluate the derivative at the two specified inputs. At x = −1, f′(−1) = −4 + 3 = −1; at x = 0, f′(0) = 3.
3. Substitute these derivative values into the required expression: f′(0) + 3f′(−1) = 3 + 3(−1) = 0, verifying the stated relation.

Q6. Differentiate f(x) = (x + 1)/x for x ≠ 0 using the quotient rule, and state the restriction on the derivative. [3 marks]

1. Set u = x + 1 for the numerator and v = x for the denominator. Their derivatives are u′ = 1 and v′ = 1.
2. Apply the quotient rule in its stated order: f′(x) = (u′v − uv′)/v² = [1 × x − (x + 1) × 1]/x².
3. Simplify the numerator to −1, giving f′(x) = −1/x². This derivative is defined for x ≠ 0, retaining the excluded input of the original function.

Q7. Differentiate f(x) = sin² x using the product rule, with x in radians. Express the answer using the identity sin(2x) = 2 sin x cos x. [2 marks]

1. Write sin² x as sin x sin x. The product rule gives f′(x) = cos x sin x + sin x cos x.
2. Combine the terms to obtain 2 sin x cos x. The supplied identity gives the equivalent answer f′(x) = sin(2x).

Q8. For a fixed base b > 0 with b ≠ 1, evaluate the limit of log to base b of (1 + x), divided by x, as x → 0. Use log to base b of z = ln z/ln b for z > 0, and ln(1 + x)/x → 1. [3 marks]

1. The logarithm is defined when 1 + x > 0. During the limiting calculation also take x ≠ 0, because the given quotient divides by x.
2. Apply the supplied change-of-base formula to rewrite the quotient as [ln(1 + x)/x]/ln b. The factor 1/ln b is constant because the base is fixed.
3. The supplied natural-logarithm quotient tends to 1. Since b ≠ 1 makes ln b non-zero, the required limit is 1/ln b.

Key takeaways

• A limit concerns nearby inputs; the value at the limiting input may be different or may be undefined.
• A finite two-sided limit requires the left-hand and right-hand limits to exist and have the same value.
• The quotient law for limits requires a non-zero denominator limit; simplify indeterminate expressions before applying it.
• Use radians for standard trigonometric limits and derivatives, and match a sine function with its angle in the denominator.
• Exponential and logarithmic limits require attention to the base, while logarithmic arguments must remain positive.
• The derivative is a limit of average rates and gives the tangent slope wherever the finite defining limit exists.
• From first principles, simplify with a non-zero increment, take its limit, and retain the original domain restrictions.
• Use sum, difference, product and quotient rules with their conditions; differentiate before evaluating at a specified point.

Test yourself

Can a function have a limit at a point where it is undefined?

Yes. For (x² − 4)/(x − 2), the limit as x → 2 is 4, although the function is undefined at 2.

What must be true before dividing two finite limits?

Both limits must exist, and the denominator's limit must be non-zero for the quotient law to apply.

What does the form 0/0 tell you about a limit?

It does not determine the answer. Rewrite the expression using a valid simplification before evaluating the limit.

What is the limit of tan x/x as x → 0 in radians?

It is 1, since tan x/x equals (sin x/x)/cos x and the two relevant limits are both 1.

Why is h kept non-zero in a first-principles calculation?

The difference quotient divides by h. Simplification takes place for non-zero h before the limit as h → 0 is taken.

How does a derivative relate to a tangent?

Where the finite derivative exists, it equals the tangent's slope, obtained as the limiting slope of nearby secants.

What is the derivative of a constant function, and why?

It is zero because the output change is zero for every non-zero increment in the input.

Why is ln(1 + x)/x different from a quotient using logarithms to another base?

The change-of-base formula introduces a constant factor 1/ln b. Consequently, the latter quotient has limit 1/ln b rather than necessarily 1.