Why does the rule need the derivative of the denominator to be non-zero near the point?

If the denominator's derivative vanishes near the point, the ratio of derivatives can itself become indeterminate or undefined, breaking the reasoning. This condition rules out pathological cases.

Can I differentiate several times before checking the form?

No. Check the form after every single differentiation. Applying L'Hospital to a form that is not indeterminate gives incorrect answers. This is one of the most common student errors.

L’Hospital’s Rule

Q: Can L'Hospital's rule be applied when the limit is at infinity?

Yes. The rule works identically when x tends to plus or minus infinity, provided the form is zero over zero or infinity over infinity and the derivative ratio has a limit.

Q: Is L'Hospital always better than series expansion?

No. For problems needing the leading power of x, expansion is faster. L'Hospital is better for one or two clean applications. For problems combining trig, exponential, and log, expansion often outperforms L'Hospital.

Q: Does L'Hospital apply to one-sided limits?

Yes. The rule works for right-hand, left-hand, and two-sided limits equally, provided the indeterminate form and differentiability conditions hold from the relevant side.

Q: What is the difference between the first form and the stronger form?

The first form assumes f and g are defined at a with both equal to zero, and uses the ratio of derivatives at a directly. The stronger form only requires limits to be zero or plus-minus infinity, does not need function values at a, and uses the limit of the derivative ratio. Always use the stronger form in practice.

Q: Does L'Hospital work for sequences?

Not directly, because n is discrete and derivatives are not defined on integers. The trick is to replace n by a continuous variable x and apply L'Hospital to the continuous limit; if that limit exists, so does the sequence limit and they agree.

MathsLimits, Continuity And DifferentiabilityFor JEE aspirants

L'Hospital's Rule states that if $lim_{x \to a} \frac{f ( x )}{g ( x )}$ is of the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , and $f$ and $g$ are differentiable near $a$ with $g^{'} (x) \neq = 0$ , then $lim_{x \to a} \frac{f ( x )}{g ( x )} = lim_{x \to a} \frac{f ^{'} ( x )}{g ^{'} ( x )}$ , provided the right-hand limit exists (finite or infinite). The rule applies at any point $a$ , at $x \to \pm \infty$ , and can be applied repeatedly as long as an indeterminate form persists. Other indeterminate forms ( $0 \cdot \infty$ , $\infty - \infty$ , $0^{0}$ , $\infty^{0}$ , $1^{\infty}$ ) must first be rewritten as $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before the rule applies. It is one of the most powerful techniques in JEE calculus.

Key Statements - Quick Reference

First form: if $f (a) = g (a) = 0$ , $f^{'} (a)$ and $g^{'} (a)$ exist, and $g^{'} (a) \neq = 0$ , then $lim_{x \to a} \frac{f ( x )}{g ( x )} = \frac{f ^{'} ( a )}{g ^{'} ( a )}$ .
Stronger form: if $lim_{x \to a} f (x) = lim_{x \to a} g (x) = 0$ (or both $= \pm \infty$ ), $f$ and $g$ are differentiable on an open interval containing $a$ (except possibly at $a$ ), and $g^{'} (x) \neq = 0$ near $a$ , then $lim_{x \to a} \frac{f ( x )}{g ( x )} = lim_{x \to a} \frac{f ^{'} ( x )}{g ^{'} ( x )}$ , provided the right-hand limit exists.
Applicable forms directly: $\frac{0}{0}$ and $\frac{\infty}{\infty}$ only.
Reducible forms: $0 \cdot \infty$ (rewrite as $\frac{0}{1/\infty}$ ), $\infty - \infty$ (combine into single fraction), $0^{0}$ , $\infty^{0}$ , $1^{\infty}$ (take logarithm first).
The rule also applies as $x \to \pm \infty$ .
Repeated application is allowed as long as the derivative ratio remains indeterminate.

1. The Statement

1.1 First Form

Suppose

f (a) = g (a) = 0

, that

f^{'} (a)

and

g^{'} (a)

both exist, and

g^{'} (a) \neq = 0

. Then

x \to a lim \frac{f ( x )}{g ( x )} = \frac{f ^{'} ( a )}{g ^{'} ( a )} .

Why it works, intuitively: near $a$ , both $f$ and $g$ are approximately linear (by their first-order Taylor expansions). So the ratio $\frac{f ( x )}{g ( x )} \approx \frac{f ^{'} ( a ) ( x - a )}{g ^{'} ( a ) ( x - a )} = \frac{f ^{'} ( a )}{g ^{'} ( a )}$ .

1.2 Stronger Form

Suppose

lim_{x \to a} f (x) = lim_{x \to a} g (x) = 0

(or both are

\pm \infty

), that

f

and

g

are differentiable on an open interval

I

containing

a

(except possibly at

a

), and

g^{'} (x) \neq = 0

I

for

x \neq = a

. Then

x \to a lim \frac{f ( x )}{g ( x )} = x \to a lim \frac{f ^{'} ( x )}{g ^{'} ( x )},

provided the right-hand limit exists (as a finite number,

+ \infty

, or

- \infty

The stronger form is what is used in practice. It does not require $f$ or $g$ to be defined at $a$ , only that both limits are $0$ (or both are $\pm \infty$ ), and it works for $x \to \pm \infty$ as well.

2. The Four Conditions - Do Not Skip Them

Before writing

lim \frac{f}{g} = lim \frac{f ^{'}}{g ^{'}}

, confirm all four:

The form is $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .
$f$ and $g$ are differentiable in a neighbourhood of $a$ (except possibly at $a$ ).
$g^{'} (x) \neq = 0$ near $a$ (except possibly at $a$ ).
$lim_{x \to a} \frac{f ^{'} ( x )}{g ^{'} ( x )}$ exists (finite or $\pm \infty$ ).

Skipping condition 1 is the most common student error - applying L'Hospital to a form like $\frac{sin x}{x + 1}$ at $x = 0$ (which is $\frac{0}{1} = 0$ directly) gives the wrong answer. Skipping condition 4 leads to loops or "the limit does not exist" false conclusions.

3. The Decision Flowchart

Figure 1: Decision flow for applying L'Hospital's rule. Repeat the derivative step as long as an indeterminate form persists; stop and evaluate as soon as a definite value appears.

4. Standard Applications: $\frac{0}{0}$ and $\frac{\infty}{\infty}$

Solved Example 1

Evaluate

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

Solution:

Form $\frac{0}{0}$ . Apply L'Hospital by differentiating with respect to $h$ (treat $x$ as a constant):

h \to 0 lim \frac{\frac{d}{d h} ( x + h - x )}{\frac{d}{d h} ( h )} = h \to 0 lim \frac{\frac{1}{2 x + h}}{1} = \frac{1}{2 x} .

This is exactly the derivative of $x$ from first principles - a reassuring check.

Solved Example 2

Evaluate

lim_{x \to 0} \frac{e ^{x} - 1 - x}{x ^{2}}

Solution:

Form $\frac{0}{0}$ . First application:

x \to 0 lim \frac{e ^{x} - 1}{2 x} .

Still $\frac{0}{0}$ . Second application:

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

Solved Example 3

Evaluate

lim_{x \to \infty} \frac{x ^{n}}{e ^{x}}

for any positive integer

n

Solution:

Form $\frac{\infty}{\infty}$ . Apply L'Hospital $n$ times:

x \to \infty lim \frac{x ^{n}}{e ^{x}} = x \to \infty lim \frac{n x ^{n - 1}}{e ^{x}} = x \to \infty lim \frac{n ( n - 1 ) x ^{n - 2}}{e ^{x}} = \dots = x \to \infty lim \frac{n !}{e ^{x}} = 0.

This confirms the standard fact: exponential growth beats every polynomial.

Solved Example 4

Evaluate

lim_{x \to 1} \frac{1 + ln x - x}{1 - 2 x + x ^{2}}

Solution:

At $x = 1$ : numerator $= 1 + 0 - 1 = 0$ ; denominator $= (x - 1)^{2} = 0$ . Form $\frac{0}{0}$ .

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

5. Reducing Other Indeterminate Forms

L'Hospital only handles $\frac{0}{0}$ and $\frac{\infty}{\infty}$ directly. Other forms must first be rewritten.

5.1 The form $0 \cdot \infty$

Rewrite $f \cdot g$ as $\frac{f}{1/ g}$ (giving $\frac{0}{0}$ ) or $\frac{g}{1/ f}$ (giving $\frac{\infty}{\infty}$ ). Choose whichever gives easier derivatives.

Solved Example 5

Evaluate

lim_{x \to π / 2^{-}} tan x \cdot ln (sin x)

Solution:

Form $\infty \cdot 0$ (since $tan x \to \infty$ and $ln (sin x) \to ln 1 = 0$ ). Rewrite:

x \to π / 2^{-} lim \frac{ln ( sin x )}{cot x} (form \frac{0}{0}) .

Apply L'Hospital:

x \to π / 2^{-} lim \frac{\frac{c o s x}{s i n x}}{- csc ^{2} x} = x \to π / 2^{-} lim \frac{cos x}{sin x} \cdot (- sin^{2} x) = x \to π / 2^{-} lim (- cos x sin x) = 0.

5.2 The form $\infty - \infty$

Combine the two terms into a single fraction (common denominator, rationalization, or factoring the dominant term).

5.3 The forms $0^{0}$ , $\infty^{0}$ , $1^{\infty}$

Take natural log first: if $L = lim f^{g}$ , then $ln L = lim g ln f$ , which typically converts to $0 \cdot \infty$ or $\infty \cdot 0$ ; then rewrite as a fraction and apply L'Hospital. Finally, $L = e^{l n L}$ .

Solved Example 6

Evaluate

lim_{x \to \infty} x^{1/ x}

Solution:

Form $\infty^{0}$ . Let $L = lim_{x \to \infty} x^{1/ x}$ . Then

ln L = x \to \infty lim \frac{ln x}{x} (form \frac{\infty}{\infty}) .

Apply L'Hospital:

ln L = x \to \infty lim \frac{1/ x}{1} = 0 ⟹ L = e^{0} = 1.

6. A Delicate Case: The Integral Form

Solved Example 7

Evaluate

lim_{x \to 0} \frac{x - \int _{0}^{x} cos ( t ^{2} ) d t}{x ^{3} - 6 x}

Solution:

At $x = 0$ : numerator $= 0 - 0 = 0$ ; denominator $= 0$ . Form $\frac{0}{0}$ . Differentiate using the fundamental theorem of calculus for the integral:

\frac{d}{d x} (x - \int_{0}^{x} cos (t^{2}) d t) = 1 - cos (x^{2}),

\frac{d}{d x} (x^{3} - 6 x) = 3 x^{2} - 6.

x \to 0 lim \frac{1 - cos ( x ^{2} )}{3 x ^{2} - 6} = \frac{1 - 1}{0 - 6} = 0.

7. When L'Hospital Does Not Help

L'Hospital can fail in three ways:

7.1 The derivative ratio has no limit

Consider $lim_{x \to \infty} \frac{x + sin x}{x}$ . Direct calculation: this is $lim_{x \to \infty} (1 + \frac{s i n x}{x}) = 1$ . But applying L'Hospital gives $lim_{x \to \infty} \frac{1 + cos x}{1}$ , which oscillates and has no limit. The rule breaks down because condition 4 (derivative limit exists) fails - yet the original limit exists. Always evaluate the derivative ratio; if it does not exist, the rule tells you nothing about the original limit, and you should use another method.

7.2 The process loops

Some limits regenerate the same form after each application. Example: $lim_{x \to \infty} \frac{e ^{x} + e ^{- x}}{e ^{x} - e ^{- x}}$ keeps returning to itself. Switch to algebraic manipulation: divide numerator and denominator by $e^{x}$ to get $\frac{1 + e ^{- 2 x}}{1 - e ^{- 2 x}} \to 1$ .

7.3 The form is not indeterminate

Applying L'Hospital to a limit that already has a definite value produces the wrong answer. Always check the form before differentiating.

8. L'Hospital vs Series Expansion - When to Use Which

Situation	Prefer
Simple $\frac{0}{0}$ that resolves in 1-2 applications	L'Hospital
Repeated applications get messy or hard to differentiate	Series expansion
Problem asks "find $n$ so the limit is finite non-zero"	Series expansion
Numerator and denominator both algebraic and factorable	Factorization first, then L'Hospital if needed
Trig, exponential, or log at $0$ with a clean standard form	Standard limits first
Integral in the numerator or denominator	L'Hospital (with fundamental theorem)

Common Mistakes to Avoid

Watch out

Applying L'Hospital without checking the form. For $lim_{x \to 0} \frac{sin x}{x + 1}$ , direct substitution gives $\frac{0}{1} = 0$ . Applying L'Hospital gives $lim_{x \to 0} cos x = 1$ , which is wrong.
Differentiating using the quotient rule. L'Hospital differentiates numerator and denominator separately. The quotient rule is not the same and will not give the right answer.
Concluding the limit does not exist when only the derivative ratio has no limit. Failure of condition 4 means the rule is silent, not that the original limit is undefined.
Endless application without checking. After each application, check whether an indeterminate form still exists. If not, stop and evaluate.
Forgetting to convert other forms first. L'Hospital does not apply directly to $0 \cdot \infty$ , $\infty - \infty$ , or exponential indeterminate forms. Convert to $\frac{0}{0}$ or $\frac{\infty}{\infty}$ first.
Missing hidden zero derivatives in the denominator. Condition 3 requires $g^{'} (x) \neq = 0$ near $a$ . Oscillatory denominators (like $x + sin x$ ) may violate this near $x = 0$ .

Frequently Asked Questions

Q1. Can L'Hospital's rule be applied when the limit is at infinity?

Yes. The rule works identically when $x \to + \infty$ or $x \to - \infty$ , provided the form is $\frac{0}{0}$ or $\frac{\infty}{\infty}$ and the derivative ratio has a limit. This is why we can prove $lim_{x \to \infty} \frac{x ^{n}}{e ^{x}} = 0$ by applying it $n$ times.

Q2. How many times can I apply L'Hospital in one problem?

As many times as needed, provided the form remains indeterminate ( $\frac{0}{0}$ or $\frac{\infty}{\infty}$ ) and the differentiability conditions continue to hold. After each application, re-check the form. Once you get a definite value, stop and report it.

Q3. Is L'Hospital always better than series expansion?

No. For problems that need identification of the leading order of $x$ , expansion is faster and clearer. L'Hospital is faster for one or two clean applications. For problems combining trig, exponential, and log all together, expansion often outperforms L'Hospital because differentiating repeatedly can produce enormous expressions.

Q4. Why does the rule need $g^{'} (x) \neq = 0$ near $a$ ?

If $g^{'}$ vanishes near $a$ , the ratio $\frac{f ^{'} ( x )}{g ^{'} ( x )}$ can itself be an indeterminate form or undefined, breaking the reasoning. In practice, this condition rules out pathological cases like oscillating derivatives; most standard JEE problems satisfy it without extra work.

Q5. Does L'Hospital apply to one-sided limits?

Yes. The rule works for $x \to a^{+}$ , $x \to a^{-}$ , and two-sided $x \to a$ equally, provided the indeterminate form and differentiability conditions hold from the relevant side.

Q6. What is the difference between the first form and the stronger form?

The first form assumes $f$ and $g$ are actually defined at $a$ with $f (a) = g (a) = 0$ and uses $\frac{f ^{'} ( a )}{g ^{'} ( a )}$ directly. The stronger form only requires the limits to be $0$ (or $\pm \infty$ ), does not need $f (a)$ or $g (a)$ to exist, and computes $lim_{x \to a} \frac{f ^{'} ( x )}{g ^{'} ( x )}$ . In practice, always use the stronger form.

Q7. Can I differentiate the numerator and denominator any number of times before checking?

No. Check the form after every single differentiation. Applying L'Hospital to a non-indeterminate form gives incorrect answers. This is one of the most common errors: mechanically differentiating three or four times without pausing to see whether the form already resolved.

Q8. Does L'Hospital work for sequences (limits as $n \to \infty$ over integers)?

Not directly, because $n$ is a discrete variable and derivatives are not defined. The standard trick is to replace $n$ by a continuous variable $x$ and apply L'Hospital to the continuous limit; if the continuous limit exists, so does the sequence limit and they agree.

Previous year questions on L’Hospital’s Rule

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

Ready to master Limits, Continuity And Differentiability?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test

1. The Statement

1.1 First Form

1.2 Stronger Form

2. The Four Conditions - Do Not Skip Them

3. The Decision Flowchart

4. Standard Applications: 00​ and ∞∞​

5. Reducing Other Indeterminate Forms

5.1 The form 0⋅∞

5.2 The form ∞−∞

5.3 The forms 00, ∞0, 1∞

6. A Delicate Case: The Integral Form

7. When L'Hospital Does Not Help

7.1 The derivative ratio has no limit

7.2 The process loops

7.3 The form is not indeterminate

8. L'Hospital vs Series Expansion - When to Use Which

Common Mistakes to Avoid

Frequently Asked Questions

Q1. Can L'Hospital's rule be applied when the limit is at infinity?

Q2. How many times can I apply L'Hospital in one problem?

Q3. Is L'Hospital always better than series expansion?

Q4. Why does the rule need g′(x)=0 near a?

Q5. Does L'Hospital apply to one-sided limits?

Q6. What is the difference between the first form and the stronger form?

Q7. Can I differentiate the numerator and denominator any number of times before checking?

Q8. Does L'Hospital work for sequences (limits as n→∞ over integers)?

Previous year questions on L’Hospital’s Rule

Ready to master Limits, Continuity And Differentiability?

4. Standard Applications: $\frac{0}{0}$ and $\frac{\infty}{\infty}$

5.1 The form $0 \cdot \infty$

5.2 The form $\infty - \infty$

5.3 The forms $0^{0}$ , $\infty^{0}$ , $1^{\infty}$

Q4. Why does the rule need $g^{'} (x) \neq = 0$ near $a$ ?

Q8. Does L'Hospital work for sequences (limits as $n \to \infty$ over integers)?