Which method should I try first for evaluating a limit?

Always try direct substitution first. If it gives a finite number, that is the answer. If it gives an indeterminate form, pick the method based on the functions present: factorize polynomials, rationalize surds, use standard limits for trig and exponential and logarithmic near zero, use logarithm method for zero-power-zero and infinity-power-zero forms.

How do I recognise a one-power-infinity form?

You get one-power-infinity when the base tends to one and the exponent tends to infinity. Classic patterns include one plus one over x, all raised to the power x, or a polynomial divided by itself raised to a blowing-up exponent.

Can I mix and match methods in one problem?

Yes, and often you must. A typical problem uses rationalization to clear a surd, then factoring, then a standard trigonometric limit. Break the expression into pieces, use the algebra of limits on each, then multiply.

Why does the limit of (a to the x minus 1) over x equal ln a?

Write a to the x as e to the x ln a. Then the expression becomes (e to the y minus 1) over y, all multiplied by ln a, with y equal to x ln a. As x tends to zero, y tends to zero and the first factor tends to one, so the whole expression tends to ln a.

Methods to Evaluate Limits

Q: When should I use series expansion instead of L'Hospital's rule?

Series expansion is faster when both numerator and denominator need many differentiations, or when you need to identify the leading power of x. L'Hospital is quicker for one or two applications on clean zero-over-zero or infinity-over-infinity forms. For problems asking for the value of n that makes the limit finite non-zero, expansion is preferred.

Q: Do standard trigonometric limits require x to be in radians?

Yes. The standard limit sin x over x tends to 1 assumes x is in radians. In degrees, sin x near zero behaves like pi x over 180, so the limit becomes pi over 180, not 1. All calculus uses radians unless stated otherwise.

Q: What is the role of the sandwich theorem?

The sandwich theorem is essential for limits involving the greatest integer function, oscillatory factors like sine of one over x, and sums involving greatest integer expressions. Whenever direct techniques fail because the function is not smooth, sandwich it between two smoother functions with the same limit.

Q: How many terms of a Taylor series should I keep?

Keep enough terms so that after all cancellations, the surviving lowest power in numerator and denominator are both accounted for. If the denominator is x cubed, retain terms up to x cubed in the numerator. When in doubt, keep one extra term and drop it at the end.

MathsLimits, Continuity And DifferentiabilityFor JEE aspirants

When direct substitution into $lim_{x \to a} f (x)$ gives an indeterminate form, we resolve it using one of six standard methods: factorization, rationalization, substitution (change of variable), series expansion, application of standard limits, and logarithm-based methods for $0^{0}$ , $\infty^{0}$ , and $1^{\infty}$ forms. The right method depends on the form: $0/0$ with polynomials calls for factoring, surds call for rationalizing, and forms involving $sin, tan, e^{x}, ln$ near $0$ call for standard limits. This concept is the workhorse of JEE calculus and appears in almost every calculus problem downstream.

Standard Limits - Quick Reference

$lim_{x \to a} \frac{x ^{n} - a ^{n}}{x - a} = n a^{n - 1}$ .
$lim_{x \to 0} \frac{sin x}{x} = 1 = lim_{x \to 0} \frac{tan x}{x} = lim_{x \to 0} \frac{sin ^{- 1} x}{x} = lim_{x \to 0} \frac{tan ^{- 1} x}{x}$ (with $x$ in radians).
$lim_{x \to 0} \frac{a ^{x} - 1}{x} = ln a$ for $a > 0$ ; in particular $lim_{x \to 0} \frac{e ^{x} - 1}{x} = 1$ .
$lim_{x \to 0} \frac{ln ( 1 + x )}{x} = 1$ and $lim_{x \to 0} \frac{lo g _{a} ( 1 + x )}{x} = lo g_{a} e$ .
$lim_{x \to 0} (1 + x)^{1/ x} = e$ ; equivalently $lim_{x \to \infty} (1 + \frac{1}{x})^{x} = e$ and $lim_{x \to \infty} (1 + \frac{a}{x})^{x} = e^{a}$ .
For $1^{\infty}$ : if $lim f (x) = 1$ and $lim ϕ (x) = \infty$ , then $lim f (x)^{ϕ (x)} = e^{l i m (f (x) - 1) ϕ (x)}$ .
Sandwich theorem: if $f (x) \leq g (x) \leq h (x)$ near $a$ and $lim_{x \to a} f = lim_{x \to a} h = ℓ$ , then $lim_{x \to a} g = ℓ$ .

1. Method 1: Direct Substitution

Always try this first. If substituting $x = a$ into $f (x)$ gives a finite number, that number is the limit. This works because polynomials, rational functions (away from zeros of the denominator), $sin, cos, exp$ , and $ln$ (in its domain) are all continuous.

Solved Example 1

Evaluate

lim_{x \to 2} \frac{x ^{2} - 4}{x + 3}

Solution:

Substitute $x = 2$ directly: numerator $= 0$ , denominator $= 5$ , both finite. So

x \to 2 lim \frac{x ^{2} - 4}{x + 3} = \frac{0}{5} = 0.

If substitution gives an indeterminate form, move on to one of the methods below.

2. Method 2: Factorization

Use when $lim_{x \to a} \frac{f ( x )}{g ( x )}$ gives $\frac{0}{0}$ and both $f$ and $g$ are polynomials (or factor cleanly). Since $f (a) = g (a) = 0$ , both have $(x - a)$ as a factor. Factor it out from numerator and denominator, cancel, and re-evaluate. Repeat if needed.

Useful factoring identities:

Factoring identities

$x^{n} - a^{n} = (x - a) (x^{n - 1} + x^{n - 2} a + \dots + a^{n - 1})$ , giving $lim_{x \to a} \frac{x ^{n} - a ^{n}}{x - a} = n a^{n - 1}$ .
$(1 + x)^{n} - 1 = x \cdot n + (2 n) x^{2} + \dots$ , giving $lim_{x \to 0} \frac{( 1 + x ) ^{n} - 1}{x} = n$ .

Solved Example 2

Evaluate

lim_{x \to 2} \frac{x ^{2} - 5 x + 6}{x ^{2} - 4}

Solution:

Direct substitution gives $\frac{0}{0}$ . Factor both:

x \to 2 lim \frac{( x - 2 ) ( x - 3 )}{( x - 2 ) ( x + 2 )} = x \to 2 lim \frac{x - 3}{x + 2} = \frac{- 1}{4} .

The cancellation is legal because $x \to 2$ means $x \neq = 2$ , so $x - 2 \neq = 0$ .

Solved Example 3

Evaluate

lim_{x \to 1} \frac{x ^{3} - 1}{x - 1}

Solution:

Form $\frac{0}{0}$ . Use $x^{3} - 1 = (x - 1) (x^{2} + x + 1)$ :

x \to 1 lim \frac{( x - 1 ) ( x ^{2} + x + 1 )}{x - 1} = x \to 1 lim (x^{2} + x + 1) = 3.

3. Method 3: Rationalization

Use when the expression contains square roots (or higher roots) and direct substitution gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . Multiply numerator and denominator by the conjugate of the surd expression to clear the root.

Solved Example 4

Evaluate

lim_{x \to 0} \frac{2 + x - 2}{x}

Solution:

Form $\frac{0}{0}$ . Multiply top and bottom by the conjugate $2 + x + 2$ :

x \to 0 lim \frac{( 2 + x - 2 ) ( 2 + x + 2 )}{x ( 2 + x + 2 )} = x \to 0 lim \frac{( 2 + x ) - 2}{x ( 2 + x + 2 )}

= x \to 0 lim \frac{1}{2 + x + 2} = \frac{1}{2 2} .

Solved Example 5

Evaluate

lim_{x \to - \infty} (25 x^{2} - 3 x + 5 x)

Solution:

Form $\infty - \infty$ . Substitute $x = - y$ so $y \to \infty$ :

y \to \infty lim (25 y^{2} + 3 y - 5 y) .

Now rationalize by multiplying with the conjugate $25 y^{2} + 3 y + 5 y$ :

= y \to \infty lim \frac{( 25 y ^{2} + 3 y ) - 25 y ^{2}}{25 y ^{2} + 3 y + 5 y} = y \to \infty lim \frac{3 y}{25 y ^{2} + 3 y + 5 y} .

Divide top and bottom by $y$ :

= y \to \infty lim \frac{3}{25 + 3/ y + 5} = \frac{3}{5 + 5} = \frac{3}{10} .

4. Method 4: Standard Trigonometric Limits

For

x

in radians:

x \to 0 lim \frac{sin x}{x} = 1, x \to 0 lim \frac{tan x}{x} = 1, x \to 0 lim \frac{sin ^{- 1} x}{x} = 1, x \to 0 lim \frac{tan ^{- 1} x}{x} = 1.

Also useful:

lim_{x \to \infty} \frac{sin x}{x} = 0

(since

∣ sin x ∣ \leq 1

Argument-matching trick. For

lim_{x \to 0} \frac{sin ( k x )}{x}

, multiply and divide by

k

\frac{sin ( k x )}{x} = k \cdot \frac{sin ( k x )}{k x}

, and

\frac{sin ( k x )}{k x} \to 1

x \to 0

. So

lim_{x \to 0} \frac{sin ( k x )}{x} = k

Solved Example 6

Evaluate

lim_{x \to 0} \frac{sin ^{2} 2 x}{sin ^{2} 3 x}

Solution:

Rewrite each factor to expose the standard form:

\frac{sin ^{2} 2 x}{sin ^{2} 3 x} = \frac{( \frac{sin 2 x}{2 x} ) ^{2} ( 2 x ) ^{2}}{( \frac{sin 3 x}{3 x} ) ^{2} ( 3 x ) ^{2}} = \frac{( 2 x ) ^{2}}{( 3 x ) ^{2}} \cdot \frac{( \frac{sin 2 x}{2 x} ) ^{2}}{( \frac{sin 3 x}{3 x} ) ^{2}} .

As $x \to 0$ , both $sin 2 x /2 x$ and $sin 3 x /3 x$ tend to $1$ , so

x \to 0 lim \frac{sin ^{2} 2 x}{sin ^{2} 3 x} = \frac{4}{9} .

Solved Example 7

Evaluate

lim_{x \to 0} \frac{tan x - sin x}{x ^{3}}

Solution:

Write $tan x = sin x / cos x$ and simplify:

\frac{tan x - sin x}{x ^{3}} = \frac{sin x - sin x cos x}{x ^{3} cos x} = \frac{sin x ( 1 - cos x )}{x ^{3} cos x} .

Use $1 - cos x = 2 sin^{2} (x /2)$ :

= \frac{sin x}{x} \cdot \frac{2 sin ^{2} ( x /2 )}{x ^{2}} \cdot \frac{1}{cos x} = \frac{sin x}{x} \cdot \frac{1}{2} (\frac{sin ( x /2 )}{x /2})^{2} \cdot \frac{1}{cos x} .

Taking $x \to 0$ : each of $sin x / x$ and $sin (x /2) / (x /2)$ approaches $1$ , and $cos x \to 1$ , so

x \to 0 lim \frac{tan x - sin x}{x ^{3}} = 1 \cdot \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2} .

5. Method 5: Standard Exponential Limits

x \to 0 lim \frac{a ^{x} - 1}{x} = ln a (a > 0), x \to 0 lim \frac{e ^{x} - 1}{x} = 1.

Solved Example 8

Evaluate

lim_{x \to 0} \frac{a ^{x} - b ^{x}}{x}

Solution:

Add and subtract $1$ in the numerator:

x \to 0 lim \frac{( a ^{x} - 1 ) - ( b ^{x} - 1 )}{x} = x \to 0 lim \frac{a ^{x} - 1}{x} - x \to 0 lim \frac{b ^{x} - 1}{x} = ln a - ln b = ln \frac{a}{b} .

Solved Example 9

Evaluate

lim_{x \to 0} \frac{e ^{- x} - 1}{x}

Solution:

Put $- x = y$ , so $y \to 0$ as $x \to 0$ :

x \to 0 lim \frac{e ^{- x} - 1}{x} = y \to 0 lim \frac{e ^{y} - 1}{- y} = - y \to 0 lim \frac{e ^{y} - 1}{y} = - 1.

6. Method 6: Standard Logarithmic Limits

x \to 0 lim \frac{ln ( 1 + x )}{x} = 1, x \to 0 lim \frac{lo g _{a} ( 1 + x )}{x} = lo g_{a} e .

Solved Example 10

Evaluate

lim_{x \to e} \frac{ln x - 1}{x - e}

Solution:

Substitute $x = e + h$ so $h \to 0$ :

h \to 0 lim \frac{ln ( e + h ) - 1}{h} = h \to 0 lim \frac{ln ( e + h ) - ln e}{h} = h \to 0 lim \frac{ln ( 1 + \frac{h}{e} )}{h} .

Multiply and divide by $\frac{1}{e}$ :

= h \to 0 lim \frac{ln ( 1 + \frac{h}{e} )}{h / e} \cdot \frac{1}{e} = 1 \cdot \frac{1}{e} = \frac{1}{e} .

7. Method 7: Substitution (Change of Variable)

If the limit variable is not tending to $0$ but the useful standard limits require the argument to tend to $0$ , substitute $x = a + h$ (or similar) so the new variable $h \to 0$ . If the argument is trigonometric, substitute so the trig function itself becomes the new variable.

Solved Example 11

Evaluate

lim_{x \to 1} (1 - x) tan \frac{π x}{2}

Solution:

This is a $0 \times \infty$ form. Let $x - 1 = h$ , so $h \to 0$ :

h \to 0 lim (- h) tan (\frac{π}{2} + \frac{π h}{2}) = h \to 0 lim (- h) (- cot \frac{π h}{2}) = h \to 0 lim h cot \frac{π h}{2} .

Rewrite as $\frac{h}{tan ( π h /2 )}$ and multiply/divide by $π /2$ :

= h \to 0 lim \frac{π h /2}{tan ( π h /2 )} \cdot \frac{2}{π} = 1 \cdot \frac{2}{π} = \frac{2}{π} .

8. Method 8: Series Expansion (Taylor / Maclaurin)

For fine-grained comparison of small differences (typically when several standard limits together are not enough), replace each function by its Maclaurin series and keep terms up to the required order.

Essential Maclaurin expansions

$e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots$
$sin x = x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \dots$
$cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots$
$ln (1 + x) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots$ for $- 1 < x \leq 1$ .
$(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \dots$ (rational or integer $n$ ).
$a^{x} = 1 + (x ln a) + \frac{( x ln a ) ^{2}}{2 !} + \dots$
$tan x = x + \frac{x ^{3}}{3} + \frac{2 x ^{5}}{15} + \dots$
$tan^{- 1} x = x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} - \dots$

Solved Example 12

Find

a

and

b

such that

lim_{x \to 0} \frac{a e ^{x} - b}{x} = 2

Solution:

Expand $e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \dots$ :

x \to 0 lim \frac{a ( 1 + x + \frac{x ^{2}}{2 !} + \dots ) - b}{x} = x \to 0 lim \frac{( a - b ) + a x + \frac{a x ^{2}}{2 !} + \dots}{x} = 2.

For the limit to be finite, the constant term must vanish: $a - b = 0$ , so $b = a$ . Then

x \to 0 lim (a + \frac{a x}{2 !} + \dots) = a = 2.

Therefore $a = 2$ and $b = 2$ .

Solved Example 13

Find the positive integer

n

for which

lim_{x \to 0} \frac{cos ^{2} x - cos x - e ^{x} cos x + e ^{x} - \frac{x ^{3}}{2}}{x ^{n}}

is finite and non-zero.

Solution:

Group the numerator as $(cos x - 1) (cos x - e^{x}) - \frac{x ^{3}}{2}$ . Expand:

$cos x - 1 = - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots$

$cos x - e^{x} = (1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots) - (1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots) = - x - x^{2} - \frac{x ^{3}}{3 !} - \dots$

So $(cos x - 1) (cos x - e^{x}) = (- \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots) (- x - x^{2} - \frac{x ^{3}}{6} - \dots)$ . Keeping terms up to $x^{5}$ :

= \frac{x ^{3}}{2} + \frac{x ^{4}}{2} + \frac{x ^{5}}{12} - \dots

Subtracting $\frac{x ^{3}}{2}$ from the numerator kills the leading $x^{3}$ term:

numerator = \frac{x ^{4}}{2} + \frac{x ^{5}}{12} - \frac{x ^{5}}{24} + \dots = \frac{x ^{4}}{2} + \frac{x ^{5}}{24} + \dots

The lowest surviving power is $x^{4}$ , so the limit is finite and non-zero when $n = 4$ .

9. Method 9: Limits of the Form $1^{\infty}$

lim_{x \to a} f (x) = 1

and

lim_{x \to a} ϕ (x) = \infty

, then

x \to a lim f (x)^{ϕ (x)} = exp (x \to a lim (f (x) - 1) ϕ (x)) .

Derivation sketch: $f (x)^{ϕ (x)} = exp (ϕ (x) ln f (x))$ ; near $f (x) = 1$ , $ln f (x) \approx f (x) - 1$ (from $ln (1 + u) \approx u$ ). Formally, $f (x)^{ϕ (x)} = (1 + (f (x) - 1))^{ϕ (x)}$ and use $(1 + u)^{1/ u} \to e$ .

Solved Example 14

Evaluate

lim_{x \to \infty} (\frac{x + 6}{x + 1})^{x + 4}

Solution:

As $x \to \infty$ : $\frac{x + 6}{x + 1} \to 1$ and $x + 4 \to \infty$ , so this is $1^{\infty}$ . Apply the formula:

\frac{x + 6}{x + 1} - 1 = \frac{5}{x + 1},

x \to \infty lim (\frac{x + 6}{x + 1})^{x + 4} = exp (x \to \infty lim \frac{5 ( x + 4 )}{x + 1}) = exp (5) = e^{5} .

Solved Example 15

Evaluate

lim_{x \to 0} (1 + tan^{2} x)^{1/ (2 x)}

Solution:

Form $1^{\infty}$ . Apply the formula:

exp (x \to 0 lim \frac{tan ^{2} x}{2 x}) = exp (x \to 0 lim \frac{1}{2} (\frac{tan x}{x})^{2}) = exp (\frac{1}{2}) = e^{1/2} .

10. Method 10: Logarithm Method (Forms $0^{0}$ and $\infty^{0}$ )

When the form is $0^{0}$ or $\infty^{0}$ , take the natural log of the expression, evaluate the resulting limit, then exponentiate.

lim_{x \to a} f (x)^{ϕ (x)}

exists and is positive, then

x \to a lim f (x)^{ϕ (x)} = exp (x \to a lim ϕ (x) ln f (x)) .

Solved Example 16

Evaluate

lim_{x \to 0^{+}} x^{s i n x}

Solution:

Form $0^{0}$ . Take logs:

ln L = x \to 0^{+} lim sin x \cdot ln x = x \to 0^{+} lim \frac{ln x}{1/ sin x} = x \to 0^{+} lim \frac{ln x}{csc x} .

Applying L'Hospital's rule (form $\frac{- \infty}{\infty}$ ):

ln L = x \to 0^{+} lim \frac{1/ x}{- csc x cot x} = x \to 0^{+} lim \frac{- sin ^{2} x}{x cos x} = x \to 0^{+} lim - (\frac{sin x}{x})^{2} \cdot \frac{x}{cos x} = 0.

So $L = e^{0} = 1$ .

Solved Example 17

Evaluate

lim_{n \to \infty} (π n)^{2/ n}

Solution:

Form $\infty^{0}$ . Take logs:

ln L = n \to \infty lim \frac{2 ln ( π n )}{n} .

By L'Hospital's rule:

ln L = n \to \infty lim \frac{2 \cdot \frac{1}{nπ} \cdot π}{1} = n \to \infty lim \frac{2}{n} = 0.

So $L = e^{0} = 1$ .

11. Sandwich (Squeeze) Theorem

f (x) \leq g (x) \leq h (x)

for all

x

in a neighbourhood of

a

(except possibly at

a

), and

x \to a lim f (x) = x \to a lim h (x) = ℓ,

then

lim_{x \to a} g (x) = ℓ

Figure 1: The middle curve

g

is squeezed between

f

and

h

, both of which approach

L

x = a

, forcing

g \to L

as well.

Solved Example 18

Evaluate

lim_{n \to \infty} \frac{[ x ] + [ 2 x ] + [ 3 x ] + \dots + [ n x ]}{n ^{2}}

, where

[\cdot]

is the greatest integer function.

Solution:

For any real $y$ : $y - 1 < [y] \leq y$ . Sum this from $y = k x$ , $k = 1$ to $n$ :

(x + 2 x + \dots + n x) - n < k = 1 \sum n [k x] \leq (x + 2 x + \dots + n x),

\frac{x n ( n + 1 )}{2} - n < k = 1 \sum n [k x] \leq \frac{x n ( n + 1 )}{2} .

Divide by $n^{2}$ :

\frac{x}{2} (1 + \frac{1}{n}) - \frac{1}{n} < \frac{1}{n ^{2}} k = 1 \sum n [k x] \leq \frac{x}{2} (1 + \frac{1}{n}) .

As $n \to \infty$ , both outer expressions approach $\frac{x}{2}$ . By the sandwich theorem, the middle expression also approaches $\frac{x}{2}$ .

12. Choosing the Right Method - Decision Guide

What you see after direct substitution	First method to try
A finite number	Done. That's the answer.
$\frac{0}{0}$ with polynomials	Factorization
$\frac{0}{0}$ or $\frac{\infty}{\infty}$ with surds	Rationalization
$\frac{0}{0}$ with $sin, tan, sin^{- 1}, tan^{- 1}$	Standard trigonometric limits
$\frac{0}{0}$ with $e^{x}, a^{x}$	Standard exponential limits
$\frac{0}{0}$ with $ln, lo g_{a}$	Standard logarithmic limits
Variable tends to a non-zero point, but you need a " $\to 0$ " standard limit	Substitute $x = a + h$ , so $h \to 0$
$1^{\infty}$	The $1^{\infty}$ formula
$0^{0}$ or $\infty^{0}$	Logarithm method
$\frac{\infty}{\infty}$ with polynomials as $x \to \infty$	Divide by highest power of $x$
Trapped between two bounds	Sandwich theorem
Anything above that doesn't yield	Series expansion or L'Hospital's rule

Common Mistakes to Avoid

Watch out

Applying $\frac{sin x}{x} \to 1$ when $x \neq \to 0$ . The standard limit requires the argument to approach $0$ . For $lim_{x \to π} \frac{sin x}{x}$ , direct substitution gives $\frac{0}{π} = 0$ , not $1$ .
Argument mismatch in trig limits. $\frac{sin ( 3 x )}{x}$ is not $1$ as $x \to 0$ ; it is $3$ . Only $\frac{sin ( something )}{same something}$ tends to $1$ .
Cancelling factors that are not truly common. Always factor completely before cancelling. Writing $\frac{x ^{2} - 1}{x - 1} = x$ is wrong; the correct simplification is $x + 1$ .
Treating $1^{\infty}$ as $1$ . This is an indeterminate form. Use the exponent-limit formula.
Truncating a Taylor series too early. If you need behaviour up to $x^{n}$ , keep every term up to $x^{n}$ in each expansion, not just the first term. Missing higher-order terms is the most common series-method error.
Applying algebra of limits when a limit is $\infty$ . The sum/product rules require both limits to be finite.
Sandwich theorem without both bounds converging to the same value. If $f \to ℓ_{1}$ and $h \to ℓ_{2}$ with $ℓ_{1} \neq = ℓ_{2}$ , the theorem gives no information about $g$ .

Frequently Asked Questions

Q1. Which method should I try first?

Always direct substitution first. If it gives a finite number, that's the answer. If it gives an indeterminate form, look at what functions are present: polynomials to factorize, surds to rationalize, $sin$ / $tan$ / $ln$ / $e^{x}$ near $0$ to use standard limits, powers with variable base and exponent to use logarithms or the $1^{\infty}$ formula.

Q2. When should I use series expansion instead of L'Hospital's rule?

Series expansion is faster when the numerator and denominator both need many differentiations to break the indeterminate form, or when you need to identify the leading power of $x$ that survives. L'Hospital is quicker for one or two applications on clean $0/0$ or $\infty/\infty$ forms. For JEE problems of the form "find $n$ so that the limit is finite non-zero," expansion is almost always preferred.

Q3. Do standard trigonometric limits require $x$ to be in radians?

Yes. $lim_{x \to 0} \frac{sin x}{x} = 1$ assumes $x$ is in radians. In degrees, $sin x$ near $0$ behaves like $\frac{π x}{180}$ , so the limit becomes $\frac{π}{180}$ , not $1$ . All calculus is in radians unless explicitly stated otherwise.

Q4. How do I recognise a $1^{\infty}$ form?

You get $1^{\infty}$ when the base is a function tending to $1$ and the exponent is a function tending to $\infty$ (or $- \infty$ ). Classic patterns: $(1 + \frac{1}{x})^{x}$ , $(\frac{polynomial}{same polynomial})^{blows up}$ . Whenever you spot this, apply the exponent-limit formula.

Q5. Can I mix and match methods in a single problem?

Absolutely, and often you must. A typical JEE problem might use rationalization to clear a surd, then factor, then apply a standard trig limit. Break the expression into manageable pieces, use algebra of limits to handle each, then multiply the parts.

Q6. What is the role of the sandwich theorem in JEE?

The sandwich theorem is essential for limits involving the greatest integer function $[x]$ , oscillatory factors like $sin (1/ x)$ , and sums like $\sum [k x] / n^{2}$ . Whenever direct techniques fail because the function is not smooth, sandwich it between two smoother functions with the same limit.

Q7. Why does $\frac{a ^{x} - 1}{x} \to ln a$ ?

Write $a^{x} = e^{x l n a}$ . Then $\frac{a ^{x} - 1}{x} = \frac{e ^{x l n a} - 1}{x ln a} \cdot ln a$ , and as $x \to 0$ , $\frac{e ^{y} - 1}{y} \to 1$ with $y = x ln a$ , so the whole expression tends to $ln a$ .

Q8. How many terms of a Taylor series should I keep?

Keep enough terms so that after all cancellations, the surviving lowest power in the numerator and the denominator are both accounted for. If the denominator is $x^{3}$ , you must retain terms up to $x^{3}$ in the numerator. When in doubt, keep one extra term and drop it at the end.

Previous year questions on Methods to Evaluate Limits

25 questions from past papers, each with a step-by-step solution.

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1. Method 1: Direct Substitution

2. Method 2: Factorization

3. Method 3: Rationalization

4. Method 4: Standard Trigonometric Limits

5. Method 5: Standard Exponential Limits

6. Method 6: Standard Logarithmic Limits

7. Method 7: Substitution (Change of Variable)

8. Method 8: Series Expansion (Taylor / Maclaurin)

9. Method 9: Limits of the Form 1∞

10. Method 10: Logarithm Method (Forms 00 and ∞0)

11. Sandwich (Squeeze) Theorem

12. Choosing the Right Method - Decision Guide

Common Mistakes to Avoid

Frequently Asked Questions

Q1. Which method should I try first?

Q2. When should I use series expansion instead of L'Hospital's rule?

Q3. Do standard trigonometric limits require x to be in radians?

Q4. How do I recognise a 1∞ form?

Q5. Can I mix and match methods in a single problem?

Q6. What is the role of the sandwich theorem in JEE?

Q7. Why does xax−1​→lna?

Q8. How many terms of a Taylor series should I keep?

Previous year questions on Methods to Evaluate Limits

Ready to master Limits, Continuity And Differentiability?

9. Method 9: Limits of the Form $1^{\infty}$

10. Method 10: Logarithm Method (Forms $0^{0}$ and $\infty^{0}$ )

Q3. Do standard trigonometric limits require $x$ to be in radians?

Q4. How do I recognise a $1^{\infty}$ form?

Q7. Why does $\frac{a ^{x} - 1}{x} \to ln a$ ?