2. Calculus — Differential Calculus

L'Hospital's rule

L'Hospital's rule provides a powerful algorithmic mechanism for evaluating indeterminate limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ by differentiating the numerator and denominator independently.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Cauchy's Form / L'Hospital's Rule Theorem

Let $f$ and $g$ be real-valued functions differentiable on an open interval $(a, b)$ containing $c$, except possibly at $c$ itself. If $\lim_{x \to c} f(x) = 0$ and $\lim_{x \to c} g(x) = 0$ (or both tend to $\pm\infty$), and if the limit $\lim_{x \to c} \frac{f'(x)}{g'(x)}$ exists (say equal to $L$) and $g'(x) \neq 0$ for all $x \neq c$ in a neighborhood of $c$, then $$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} = L$$.

📐 Geometric Interpretation

Geometrically, as $x$ approaches $c$, the ratio of the ordinates of two curves converging to zero (or infinity) approaches the ratio of their instantaneous rates of change, namely their derivative slopes at the tangent lines.

⚙️ Algebraic & Structural Meaning

Algebraically, it bridges infinitesimal analysis with linear approximations via derivatives, reducing complex rational or transcendental limit evaluations to successive derivative computations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Indeterminacy Form	`Point $x \to c$`	Strictly Mandatory	Applying L'Hospital's rule to determinate forms (e.g., $\frac{1}{2}$) yields catastrophically false limit values.
Differentiability in Neighborhood	`Open interval $(a, b)$ around $c$`	Strictly Mandatory	If $f'(x)$ or $g'(x)$ does not exist in a punctured neighborhood of $c$, the ratio of derivatives limit may fail to exist even if the original limit exists.
Non-zero Derivative of Denominator	`$g'(x) \neq 0$ for $x \neq c$`	Essential condition	Division by zero occurs in the derivative ratio if $g'(x) = 0$ arbitrarily close to $c$.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Successive Differentiation for Polynomials

$$\lim_{x \to c} \frac{P(x)}{Q(x)} = \lim_{x \to c} \frac{P^{(n)}(x)}{Q^{(n)}(x)}$$

Rule: For polynomials yielding $\frac{0}{0}$ up to $n$-th order, differentiate numerator and denominator simultaneously $n$ times until non-zero constant is obtained.

Application: Directly tested in UP PGT for rational algebraic expressions involving high-degree roots.

Exponential-Algebraic Dominance

$$\lim_{x \to \infty} \frac{x^n}{e^{kx}} = 0 \quad (k > 0)$$

Rule: Exponential terms dominate polynomial terms; apply L'Hospital's rule $n$ times to eliminate the polynomial numerator instantly.

Application: Frequently appears in KVS/NVS limits involving growth rates.

Logarithmic Conversion for Exponents

$$\lim f(x)^{g(x)} \implies \exp\left( \lim g(x) \ln f(x) \right)$$

Rule: Convert $0^0$, $\infty^0$, or $1^\infty$ indeterminate forms into $0 \cdot \infty$ or $\frac{0}{0}$ via natural logarithm before applying L'Hospital's rule.

Series Expansion vs. L'Hospital's Rule

$$\sin x = x - \frac{x^3}{3!} + \dots$$

Rule: When multiple derivatives are tedious, replace transcendental functions with their Maclaurin series expansions up to the leading non-zero term.

Parametric Limit Evaluation

$$\lim_{x \to 0} \frac{\tan x - x}{x^3} = \frac{1}{3}$$

Rule: Memorize standard standard limits like $\frac{x - \sin x}{x^3} = \frac{1}{6}$ to bypass triple differentiation.

Option Elimination via Parameter Testing

$$a = ?, b = ?$$

Rule: Substitute convenient test values or boundary values into options to eliminate incorrect choices before full derivation.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Blind Application on Determinate Forms

Common Pitfall: Applying L'Hospital's rule blindly to limits that are not indeterminate (e.g., $\lim_{x \to 0} \frac{\sin x}{x + 1} = \frac{0}{1} = 0$).

Counterexample: $$\lim_{x \to 0} \frac{\cos x}{1} = 1 \quad (\text{Incorrect, actual limit is } 0)$$

Why it fails: The initial limit is already determinate ($0/1 = 0$), and derivative ratio gives $1/1 = 1$, which is completely false.

🛡️ Exam Defense: Always evaluate the limit by direct substitution first to confirm $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form.

⚠️ COMMON ERROR #02

Ignoring Non-Existent Derivative Ratio Limits

Common Pitfall: Assuming L'Hospital's rule fails only when indeterminate forms are absent, forgetting that if $\lim \frac{f'(x)}{g'(x)}$ does not exist, the rule gives no conclusion.

Counterexample: $$\lim_{x \to \infty} \frac{x + \sin x}{x}$$

Why it fails: Derivative ratio is $\frac{1 + \cos x}{1} = 1 + \cos x$, which oscillates and has no limit as $x \to \infty$, yet original limit is $1$.

🛡️ Exam Defense: Check if $\lim \frac{f'(x)}{g'(x)}$ actually exists before declaring the limit nonexistent.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
L'Hospital's Rule	Indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, differentiable ratio	$$\lim \frac{f}{g} = \lim \frac{f'}{g'}$$	Ratio of instantaneous tangent slopes	Ratio of functions approaching $0$ or $\infty$
Maclaurin Series	Analytic function in neighborhood of $0$	$$f(x) = \sum \frac{f^{(n)}(0)}{n!} x^n$$	Polynomial approximation curve matching derivatives at origin	Trigonometric and exponential expressions in numerator/denominator
Sandwich (Squeeze) Theorem	$g(x) \le f(x) \le h(x)$ in neighborhood	$$\lim g = \lim h = L \implies \lim f = L$$	Function squeezed between two bounding curves	Limits involving bounded functions like $\sin(1/x)$ or $\cos x$
Stolz-Cesaro Theorem	Discrete sequences, analogue of L'Hospital	$$\lim \frac{a_n}{b_n} = \lim \frac{a_n - a_{n-1}}{b_n - b_{n-1}}$$	Discrete slope comparison for sequence growth	Limits of sequence ratios with divergent denominators

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Evaluate the limit: $$\lim_{x \to 0} \frac{x \cos x - \sin x}{x^3}$$

(A) $-\frac{1}{3}$(B) $-\frac{1}{6}$(C) $\frac{1}{3}$(D) $\frac{1}{6}$

📚 Method A: Formal Textbook Derivation

Step 1: Check form: numerator $\to 0$, denominator $\to 0$ (Form $\frac{0}{0}$). Step 2: Apply L'Hospital's rule: derivative of numerator is $-x \sin x$, denominator is $3x^2$. Still $\frac{0}{0}$. Step 3: Differentiate again: numerator $-(\sin x + x \cos x)$, denominator $6x$. Still $\frac{0}{0}$. Step 4: Differentiate third time: numerator $- (2\cos x - x \sin x)$, denominator $6$. Step 5: Substitute $x = 0$ to get $\frac{-2}{6} = -\frac{1}{6}$.

⚡ Method B: 25-Second Exam Shortcut

Use standard expansion: $\sin x = x - \frac{x^3}{6} + \dots$, so $x \cos x - \sin x = x(1 - x^2/2) - (x - x^3/6) = -\frac{x^3}{3}$. Dividing by $x^3$ yields $-\frac{1}{3}$, wait, careful: $-x^3/2 + x^3/6 = -x^3/3$. Thus limit is $-\frac{1}{3}$? Let's check: $x(1 - x^2/2 + \dots) - (x - x^3/6 + \dots) = -x^3/2 + x^3/6 = -x^3/3$. Dividing by $x^3$ gives $-\frac{1}{3}$. Correct option is (A).

💡 Tip: Always cross-verify triple L'Hospital differentiation with series expansion to avoid arithmetic sign errors.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Find the value of $k$ if $$\lim_{x \to 0} \frac{e^{kx} - 1 - kx}{x^2} = 5$$

(A) $5$(B) $10$(C) $\pm 5$(D) $\pm \sqrt{10}$

📚 Method A: Formal Textbook Derivation

Step 1: Form is $\frac{0}{0}$ at $x=0$. Step 2: Apply L'Hospital's rule: $\lim_{x \to 0} \frac{k e^{kx} - k}{2x}$. Still $\frac{0}{0}$. Step 3: Apply L'Hospital's rule again: $\lim_{x \to 0} \frac{k^2 e^{kx}}{2} = \frac{k^2}{2}$. Step 4: Equate to $5$: $\frac{k^2}{2} = 5 \implies k^2 = 10 \implies k = \pm \sqrt{10}$? Wait, let's re-evaluate: series expansion for $e^{kx}$ is $1 + kx + \frac{k^2 x^2}{2} + \dots$. Subtracting $1 + kx$ leaves $\frac{k^2 x^2}{2}$. Dividing by $x^2$ gives $\frac{k^2}{2} = 5 \implies k^2 = 10 \implies k = \pm \sqrt{10}$.

⚡ Method B: 25-Second Exam Shortcut

Instantly use Taylor series quadratic coefficient: coefficient of $x^2$ in numerator is $\frac{k^2}{2}$, divided by $1$ from denominator gives $\frac{k^2}{2} = 5 \implies k = \pm \sqrt{10}$.

💡 Tip: Parameter determination problems frequently appear in UP PGT; equating Taylor coefficients is 3x faster than repeated differentiation.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: Evaluate $$\lim_{x \to 0} \frac{\tan x - x}{x - \sin x}$$

A $1$

B $2$

C $3$

D $4$

Derivation & Analysis:

Applying L'Hospital's rule three times or using series expansions $\tan x \approx x + \frac{x^3}{3}$ and $\sin x \approx x - \frac{x^3}{6}$ yields $\frac{x^3/3}{x^3/6} = 2$.

MCQ 2: The value of $$\lim_{x \to 0} \frac{\log_e(1 + x) - x}{x^2}$$ is

A $0$

B $-\frac{1}{2}$

C $1$

D $\frac{1}{2}$

Derivation & Analysis:

Form is $\frac{0}{0}$. Differentiate numerator: $\frac{1}{1+x} - 1$. Differentiate denominator: $2x$. Limit becomes $\lim_{x \to 0} \frac{-x}{(1+x)2x} = -\frac{1}{2}$.

MCQ 3: Evaluate $$\lim_{x \to 0} \frac{a^x - b^x}{x}$$

A $\log(a/b)$

B $\log(ab)$

C $\log(a^b)$

D $\log(b/a)$

Derivation & Analysis:

By L'Hospital's rule, derivative of numerator is $a^x \ln a - b^x \ln b$, denominator is $1$. At $x=0$, result is $\ln a - \ln b = \ln(a/b)$.

MCQ 4: Evaluate $$\lim_{x \to \infty} x \left( \arctan\left(\frac{x+1}{x}\right) - \frac{\pi}{4} \right)$$

A $\frac{1}{2}$

B $1$

C $-\frac{1}{2}$

D $0$

Derivation & Analysis:

Rewrite as $\frac{\arctan(1 + 1/x) - \pi/4}{1/x}$. Let $h = 1/x \to 0$. Limit becomes $\lim_{h \to 0} \frac{\arctan(1+h) - \arctan(1)}{h}$, which is the derivative of $\arctan(x)$ at $x=1$, equal to $\frac{1}{1+1^2} = \frac{1}{2}$.

MCQ 5: The limit $$\lim_{x \to 0} (1 + x)^{1/x}$$ evaluates to

A $1$

B $e$

C $e^2$

D $\infty$

Derivation & Analysis:

This is $1^\infty$ form. Let $y = (1+x)^{1/x}$, $\ln y = \frac{\ln(1+x)}{x}$. As $x \to 0$, limit is $\frac{1}{1+0} = 1$, so $y \to e^1 = e$.

MCQ 6: Evaluate $$\lim_{x \to 0} \frac{\sin(a+x) - \sin(a-x)}{x}$$

A $2 \cos a$

B $2 \sin a$

C $\cos a$

D $0$

Derivation & Analysis:

Differentiate w.r.t $x$: numerator derivative is $\cos(a+x) + \cos(a-x)$, denominator is $1$. At $x=0$, result is $2 \cos a$.

MCQ 7: If $$\lim_{x \to 0} \frac{sin 2x + a \sin x}{x^3}$$ exists finitely, then the values of $a$ and the limit are

A $a = -2$, limit $= 1$

B $a = -2$, limit $= -1$

C $a = 2$, limit $= 0$

D $a = -1$, limit $= 2$

Derivation & Analysis:

For numerator to vanish to third order, use expansion: $\sin 2x + a \sin x = (2x - \frac{8x^3}{6}) + a(x - \frac{x^3}{6}) = (2+a)x - \frac{8+a}{6}x^3$. For $x$ term to vanish, $2+a = 0 \implies a = -2$. Then coefficient of $x^3$ is $-\frac{6}{6} = -1$.

MCQ 8: Which of the following limits CANNOT be evaluated directly using L'Hospital's rule without prior algebraic transformation?

A $\lim_{x \to 0} \frac{\sin x}{x}$

B $\lim_{x \to 0^+} x \ln x$

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

D $\lim_{x \to \infty} \frac{x}{e^x}$

Derivation & Analysis:

$x \ln x$ is of form $0 \cdot \infty$, not $\frac{0}{0}$ or $\frac{\infty}{\infty}$. It must be rewritten as $\frac{\ln x}{1/x}$ or $\frac{x}{1/\ln x}$ before L'Hospital's rule applies.

MCQ 9: Evaluate $$\lim_{x \to 0} \frac{\cosh x - 1}{x^2}$$

A $1$

B $\frac{1}{2}$

C $0$

D $-\frac{1}{2}$

Derivation & Analysis:

Since $\cosh x = 1 + \frac{x^2}{2} + \dots$, the expression simplifies to $\frac{x^2/2}{x^2} = \frac{1}{2}$. Alternatively, apply L'Hospital twice: $\frac{\sinh x}{2x} \to \frac{\cosh x}{2} \to \frac{1}{2}$.

Q10

MCQ 10: Evaluate $$\lim_{x \to 0} \frac{x - \arctan x}{x^3}$$

A $\frac{1}{3}$

B $\frac{1}{6}$

C $-\frac{1}{3}$

D $1$

Derivation & Analysis:

Using series expansion $\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots$, numerator $x - (x - \frac{x^3}{3} + \dots) = \frac{x^3}{3}$. Dividing by $x^3$ gives $\frac{1}{3}$. Triple L'Hospital also yields $\frac{1}{3}$.

Common questions

Yes, provided all hypotheses (differentiability, non-zero denominator derivative) continue to hold at each stage of successive differentiation until a determinate limit is reached.

Failing to verify the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ beforehand, leading to false application on determinate values or ignoring removable discontinuities.

Master standard Maclaurin series expansions (e.g., $\sin x, \cos x, e^x, \ln(1+x)$) to bypass tedious multiple differentiations and eliminate calculation errors.