MathsCBSE Class 11 · Limits and Derivatives

Evaluating Limits Step by Step: Algebraic, Trigonometric and Exponential (Class 11)

Updated 17 October 2026 · Fact-checked

To evaluate a limit, first substitute. If the result is 0/0, remove the common factor by factorising, rationalising a square root, or rewriting with the standard limits: lim (x^n − a^n)/(x − a) = n a^(n−1), lim sin x / x = 1, lim (1 − cos x)/x² = 1/2, lim (e^x − 1)/x = 1. For a limit to exist with a zero denominator, the numerator must also vanish.

Always substitute first; only if you get 0/0 remove the zero factor by one of three tools, then substitute again.

Illustration drawn for QuestionBuddy.

Understand Evaluating Limits: Algebraic, Trigonometric and Exponential

The limit $lim_{x \to a} f (x) = L$ means $f (x)$ gets as close to $L$ as we like when $x$ is close to $a$ , whatever side it comes from (so the left and right limits must agree). Evaluation follows one routine.

Step 1: substitute. If $f (a)$ makes sense, that is the limit (for polynomials, rational functions with non-zero denominator, sine, cosine, exponentials). If you get $\frac{0}{0}$ , the factor causing the zero must be removed first.

Step 2: remove the zero.

Factorise and cancel $(x - a)$ : $\frac{x ^{3} - 8}{x ^{2} - 4} = \frac{( x - 2 ) ( x ^{2} + 2 x + 4 )}{( x - 2 ) ( x + 2 )}$ .
Rationalise when roots appear: multiply and divide by the conjugate, e.g. $\frac{1 + x - 1 - x}{x} \cdot \frac{1 + x + 1 - x}{1 + x + 1 - x}$ .
Use a standard limit, after adjusting the argument to match.

Standard limits.
$ $lim_{x \to a} \frac{x ^{n} - a ^{n}}{x - a} = n a^{n - 1}, lim_{x \to 0} \frac{s i n x}{x} = 1, lim_{x \to 0} \frac{t a n x}{x} = 1, lim_{x \to 0} \frac{1 - c o s x}{x ^{2}} = \frac{1}{2},$ $
$ $lim_{x \to 0} \frac{e ^{x} - 1}{x} = 1, lim_{x \to 0} \frac{l o g ( 1 + x )}{x} = 1.$ $
Adjusting the argument: $lim_{x \to 0} \frac{s i n k x}{x} = k$ , $lim_{x \to 0} \frac{t a n k x}{x} = k$ , $lim_{x \to 0} \frac{e ^{k x} - 1}{x} = k$ ; a ratio such as $\frac{s i n 5 x}{t a n 2 x}$ becomes $\frac{5}{2} \cdot \frac{s i n 5 x /5 x}{t a n 2 x /2 x} \to \frac{5}{2}$ . If $x \to a \neq = 0$ in a trigonometric limit, first put $x = a + h$ and let $h \to 0$ ; for instance, $\frac{1 - s i n x}{( π /2 - x ) ^{2}}$ with $h = \frac{π}{2} - x$ becomes $\frac{1 - c o s h}{h ^{2}} \to \frac{1}{2}$ .

Using the half-angle idea. $1 - cos x = 2 sin^{2} \frac{x}{2}$ and $tan x - sin x = tan x (1 - cos x)$ let you reduce to the standard limits.

Existence and constants. A limit of the form $\frac{p ( x )}{x - a}$ can be finite only if $p (a) = 0$ ; this lets you find unknown constants. For a piecewise function, the limit exists exactly when the left and right limits are equal.

Do not substitute into an indeterminate form, and do not cancel a factor that does not tend to zero. Limits are the base for differentiation, which follows on the same chapter.

Evaluating Limits: Algebraic, Trigonometric and Exponential: formulas to remember

Power limit: $l im (x^{n} - a^{n}) / (x - a) = n a^{(} n - 1) a s x \to a$; For rational n too.
Sine limit: $l im s in x / x = 1 a s x \to 0$; x in radians; sin kx / x gives k.
Cosine limit: $l im (1 - cos x) / x^{2} = 1/2 a s x \to 0$; From 1 − cos x = 2 sin²(x/2).
Exponential limit: $l im (e^{x} - 1) / x = 1 a s x \to 0$; (e^kx − 1)/x gives k.
Logarithm limit: $l im l o g (1 + x) / x = 1 a s x \to 0$; Natural logarithm.

How to solve Evaluating Limits: Algebraic, Trigonometric and Exponential questions

Use these steps for limits that start as 0/0.

1Substitute the value; if the answer is a number, stop.
2If it is $0/0$ , look for a factor to cancel (factorise), or a root to rationalise.
3For trigonometric or exponential expressions, rewrite so that the argument inside the function is the same as the denominator, then apply the standard limit.
4If $x \to a$ with $a \neq = 0$ in a trigonometric limit, put $x = a + h$ with $h \to 0$ .
5Simplify, then substitute again.
6For unknown constants, make the numerator vanish at the point first, then equate the limit to the given value.

Evaluating Limits: Algebraic, Trigonometric and Exponential in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Limit as instantaneous speed

Suppose a trolley covers $s = 4 t^{2}$ metres in $t$ seconds. To get its speed at $t = 1$ , we find its average speed over a shrinking interval from $t = 1$ to $t = 1 + h$ .

Find the average speed over that interval and the limit as $h \to 0$ .

Try it first, then show the solutionSolution

Average speed $= \frac{4 ( 1 + h ) ^{2} - 4}{h} = \frac{4 ( 2 h + h ^{2} )}{h} = 8 + 4 h$ .
As $h \to 0$ , $8 + 4 h \to 8$ .
For $h = 0.0001$ the average speed is $8.0004$ m/s, already very close.

Answer: 8 m/s

Memory hook: A limit lets h shrink to zero without ever dividing by zero.

Small-angle approximation

Suppose a pendulum engineer approximates $1 - cos θ$ by $\frac{θ ^{2}}{2}$ for small swing angles $θ$ (in radians).

Check the approximation for $θ = 0.1$ using $\frac{1 - cos θ}{θ ^{2}} \to \frac{1}{2}$ .

Try it first, then show the solutionSolution

$cos 0.1 \approx 0.995004$ , so $1 - cos 0.1 \approx 0.004996$ .
$\frac{0.004996}{0.01} \approx 0.4996$ , very close to $\frac{1}{2}$ .
Hence $1 - cos θ \approx \frac{θ ^{2}}{2}$ for small $θ$ .

Answer: The ratio is about 0.4996, close to the limit 0.5.

Memory hook: Half of theta squared: the limit of (1 − cos x)/x² is a half.

Tricks to remember Evaluating Limits: Algebraic, Trigonometric and Exponential

Short memory aids. Say each one aloud once, then use it on a question the same day.

Zero over zero means a hidden factor
The (x − a) that makes both parts vanish must be cancelled, not substituted.
Use it when: Algebraic limits.
Match the argument
Make the inside of sin, tan or e^ equal to the denominator, paying for it with a constant outside.
Use it when: Trigonometric and exponential limits.
Conjugate for roots
A difference of square roots calls for the sum of the same roots.
Use it when: Rationalising.
Shift to h
If x tends to a number other than 0, put x = a + h and let h tend to 0.
Use it when: Trigonometric limits away from 0.

Evaluating Limits: Algebraic, Trigonometric and Exponential: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Factorising · 2 marks · Very short answerBoard level

Evaluate $x \to 2 lim \frac{x ^{3} - 8}{x ^{2} - 4}$ .

Try it first, then show the solutionSolution

Substituting gives $\frac{0}{0}$ . Factorise: $\frac{( x - 2 ) ( x ^{2} + 2 x + 4 )}{( x - 2 ) ( x + 2 )}$ . (1 mark)
Cancel $(x - 2)$ and substitute: $\frac{4 + 4 + 4}{4} = 3$ . (1 mark)

Answer: 3

Example 2: Ratio of sine and tangent · 2 marks · Very short answerBoard level

Evaluate $x \to 0 lim \frac{sin 5 x}{tan 2 x}$ .

Try it first, then show the solutionSolution

Write $\frac{sin 5 x}{tan 2 x} = \frac{sin 5 x}{5 x} \cdot \frac{2 x}{tan 2 x} \cdot \frac{5}{2}$ . (1 mark)
As $x \to 0$ the first two factors tend to $1$ , so the limit is $\frac{5}{2}$ . (1 mark)

Answer: $\frac{5}{2}$

Example 3: Rationalising · 3 marks · Short answerHarder board level

Evaluate $x \to 0 lim \frac{1 + x - 1 - x}{x}$ .

Try it first, then show the solutionSolution

Multiply numerator and denominator by $1 + x + 1 - x$ . (1 mark)
Numerator becomes $(1 + x) - (1 - x) = 2 x$ , so the expression is $\frac{2 x}{x ( 1 + x + 1 - x )} = \frac{2}{1 + x + 1 - x}$ . (1 mark)
As $x \to 0$ : $\frac{2}{1 + 1} = 1$ . (1 mark)

Answer: 1

Example 4: Exponential and sine together · 3 marks · Short answerHarder board level

Evaluate $x \to 0 lim \frac{e ^{3 x} - 1}{sin 2 x}$ .

Try it first, then show the solutionSolution

Write $\frac{e ^{3 x} - 1}{sin 2 x} = \frac{e ^{3 x} - 1}{3 x} \cdot \frac{2 x}{sin 2 x} \cdot \frac{3}{2}$ . (1 mark)
First factor $\to 1$ (standard limit with $u = 3 x \to 0$ ); second factor $\to 1$ . (1 mark)
Limit $= \frac{3}{2}$ . (1 mark)

Answer: $\frac{3}{2}$

Example 5: Finding constants · 4 marks · Short answerHOTS / competency

If $x \to 1 lim \frac{x ^{2} + a x + b}{x - 1} = 5$ , find $a$ and $b$ .

Try it first, then show the solutionSolution

The denominator tends to 0 and the limit is finite, so the numerator must vanish at $x = 1$ : $1 + a + b = 0$ . (1 mark)
Then $b = - (1 + a)$ and $x^{2} + a x - (1 + a) = (x - 1) (x + 1 + a)$ . (1 mark)
Limit $= lim_{x \to 1} (x + 1 + a) = 2 + a = 5$ , so $a = 3$ . (1 mark)
$b = - 4$ . Check: $x^{2} + 3 x - 4 = (x - 1) (x + 4)$ , limit $5$ . (1 mark)

Answer: $a = 3, b = - 4$

Where students lose marks in Evaluating Limits: Algebraic, Trigonometric and Exponential

The slips examiners see most often, and what to write instead.

Mistake: Cancelling terms instead of factors
Terms joined by + or − cannot be cancelled.
Wrong working
$\frac{x ^{3} - 8}{x ^{2} - 4}$ : "cancel $x$ and $2$ ".
Right working
Factorise first, then cancel $(x - 2)$ .
Do this: Write both numerator and denominator in factored form.
Usually costs: The method marks.
Mistake: Using sin kx / x = 1
The constant k is forgotten.
Wrong working
$lim \frac{s i n 5 x}{x} = 1$ .
Right working
$\frac{s i n 5 x}{x} = 5 \cdot \frac{s i n 5 x}{5 x} \to 5$ .
Do this: Adjust the denominator to match the argument.
Usually costs: The answer.
Mistake: Degrees in standard limits
The result sin x/x → 1 holds only in radians.
Wrong working
$lim \frac{s i n x ^{\circ}}{x} = 1$ .
Right working
$\frac{s i n x ^{\circ}}{x} \to \frac{π}{180}$ .
Do this: Convert to radians first.
Usually costs: The answer.
Mistake: Not shifting x to h
Standard limits are stated at 0 only.
Wrong working
Using $\frac{1 - s i n x}{( \frac{π}{2} - x ) ^{2}}$ at $x \to \frac{π}{2}$ with $sin x / x \to 1$ .
Right working
Put $h = \frac{π}{2} - x$ , so $sin x = cos h$ and $h \to 0$ .
Do this: Whenever x tends to a non-zero number in a trig limit, use h.
Usually costs: Method and answer.

Exam tips

Always write "substituting gives 0/0" before manipulating; it justifies the method.
Show each standard limit by name when you use it (for example "using sin x / x → 1").
Keep the constant factor outside when you adjust the argument; the marks sit there.
For finding constants, state the condition on the numerator first.
Degrees and radians matter: all the standard limits use radians.

Practice questions on Evaluating Limits: Algebraic, Trigonometric and Exponential

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
$x \to 1 lim \frac{x ^{4} - 1}{x - 1}$ equals:
(a) 1
(b) 2
(c) 4
(d) 0
Hint
Use $\frac{x ^{n} - a ^{n}}{x - a} \to n a^{n - 1}$ .
Answer
(c). $4 \cdot 1^{3} = 4$ .
Q2 · 1 mark · MCQHarder board level
$x \to π /2 lim \frac{1 - sin x}{( \frac{π}{2} - x ) ^{2}}$ equals:
(a) 0
(b) $\frac{1}{2}$
(c) 1
(d) 2
Hint
Put $h = \frac{π}{2} - x$ .
Answer
(b). $sin x = cos h$ , so the limit is $lim_{h \to 0} \frac{1 - c o s h}{h ^{2}} = \frac{1}{2}$ .
Q3 · 3 marks · Short answerHarder board level
Evaluate $x \to 0 lim \frac{tan x - sin x}{x ^{3}}$ .
Hint
Write $tan x - sin x = tan x (1 - cos x)$ .
Answer
$\frac{tan x ( 1 - cos x )}{x ^{3}} = \frac{tan x}{x} \cdot \frac{1 - cos x}{x ^{2}} \to 1 \cdot \frac{1}{2} = \frac{1}{2}$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $x \to 0 lim sin \frac{1}{x}$ does not exist.
Reason (R): As $x \to 0$ , $sin \frac{1}{x}$ keeps oscillating between $- 1$ and $1$ and does not approach one value.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Think about $x = \frac{1}{nπ}$ and $x = \frac{1}{( 2 n + \frac{1}{2} ) π}$ .
Answer
(a). Along $x = \frac{1}{nπ}$ the value is $0$ ; along $x = \frac{1}{( 2 n + \frac{1}{2} ) π}$ it is $1$ . Different sequences, different values.
Q5 · 4 marks · Short answerHOTS / competency
Evaluate $x \to 0 lim \frac{1 + x sin x - 1}{x ^{2}}$ .
Hint
Rationalise, then use $\frac{s i n x}{x} \to 1$ .
Answer
Multiply by $1 + x sin x + 1$ : numerator $= x sin x$ . So the expression is $\frac{x sin x}{x ^{2} ( 1 + x sin x + 1 )} = \frac{sin x}{x} \cdot \frac{1}{1 + x sin x + 1} \to 1 \cdot \frac{1}{2} = \frac{1}{2}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is lim sin x / x as x tends to 0?Show answer
1.
2.What is lim (x^n − a^n)/(x − a)?Show answer
n a^(n−1).
3.What is lim (1 − cos x)/x²?Show answer
1/2.
4.What is lim (e^x − 1)/x?Show answer
1.
5.What must happen to the numerator if a ratio with denominator tending to zero has a finite limit?Show answer
It must also tend to zero.

Evaluating Limits: Algebraic, Trigonometric and Exponential: frequently asked questions

What does 0/0 mean for a limit?

It means the form is indeterminate. The limit may still exist, but you must simplify (factorise, rationalise or use a standard limit) before substituting.

Why do trigonometric limits use radians?

Because the standard limit sin x / x = 1 comes from the geometry of arc length, which is measured in radians.

Can a limit exist where the function is not defined?

Yes. The limit depends on values near the point, not at it.

How do I show a limit does not exist?

Show the left and right limits differ, or that the function oscillates without settling.