JEE Advanced 2003

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: $lim_{h \to 0} \frac{f ( 2 h + 2 + h ^{2} ) - f ( 2 )}{f ( h - h ^{2} + 1 ) - f ( 1 )}$ , given that $f^{'} (2) = 6$ and $f^{'} (1) = 4$

Q: Why can't I just plug in $h=0$ immediately? It gives me $0/0$, which is just 0, right?

That is the most dangerous trap in calculus! $0/0$ is not zero; it is an indeterminate form. It means the limit *might* exist, but we need to do more work to uncover it. Think of it as a locked door—plugging in $h=0$ just shows you the door is locked, but L'Hospital's Rule is the key to opening it.

Select Answer:

Visualized Solution

Analyzing the Limit Form

Given limit: $lim_{h \to 0} \frac{f ( 2 h + 2 + h ^{2} ) - f ( 2 )}{f ( h - h ^{2} + 1 ) - f ( 1 )}$
Given values: $f^{'} (2) = 6$ and $f^{'} (1) = 4$
Objective: Evaluate the limit as $h \to 0$ .

Checking Indeterminacy $\frac{0}{0}$

Substitute $h = 0$ into the expression:
Numerator: $f (2 (0) + 2 + 0^{2}) - f (2) = f (2) - f (2) = 0$
Denominator: $f (0 - 0^{2} + 1) - f (1) = f (1) - f (1) = 0$
The limit is in the indeterminate form $\frac{0}{0}$ .

Applying L'Hospital's Rule

According to L'Hospital's Rule, for $\frac{0}{0}$ forms:
$lim_{h \to 0} \frac{N ( h )}{D ( h )} = lim_{h \to 0} \frac{N ^{'} ( h )}{D ^{'} ( h )}$
We need to differentiate the numerator and denominator with respect to $h$ .

Differentiating the Numerator

Using the Chain Rule for composite functions:
$\frac{d}{d h} [f (2 h + 2 + h^{2}) - f (2)]$
$= f^{'} (2 h + 2 + h^{2}) \cdot \frac{d}{d h} (2 h + 2 + h^{2}) - 0$
$= f^{'} (2 h + 2 + h^{2}) \cdot (2 + 2 h)$

Differentiating the Denominator

Applying the Chain Rule to the denominator:
$\frac{d}{d h} [f (h - h^{2} + 1) - f (1)]$
$= f^{'} (h - h^{2} + 1) \cdot \frac{d}{d h} (h - h^{2} + 1) - 0$
$= f^{'} (h - h^{2} + 1) \cdot (1 - 2 h)$

Substituting $h = 0$

New limit expression:
$lim_{h \to 0} \frac{f ^{'} ( 2 h + 2 + h ^{2} ) \cdot ( 2 + 2 h )}{f ^{'} ( h - h ^{2} + 1 ) \cdot ( 1 - 2 h )}$
Substitute $h = 0$ :
Numerator: $f^{'} (0 + 2 + 0) \cdot (2 + 0) = f^{'} (2) \cdot 2$
Denominator: $f^{'} (0 - 0 + 1) \cdot (1 - 0) = f^{'} (1) \cdot 1$

Using Given Values

Substitute the given values $f^{'} (2) = 6$ and $f^{'} (1) = 4$ :
Numerator becomes: $6 \cdot 2 = 12$
Denominator becomes: $4 \cdot 1 = 4$
Expression becomes: $\frac{12}{4}$

Final Calculation

Calculate the final value:
$\frac{12}{4} = 3$
The limit is equal to $3$ .

00:00 / 00:00

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

We are tasked with evaluating the following limit:

h \to 0 lim \frac{f ( 2 h + 2 + h ^{2} ) - f ( 2 )}{f ( h - h ^{2} + 1 ) - f ( 1 )}

We are provided with the essential derivatives:

f^{'} (2) = 6

and

f^{'} (1) = 4

The Diagnostic

Before proceeding, we must verify the nature of the limit by direct substitution as

h \to 0

The numerator becomes

f (2 (0) + 2 + 0^{2}) - f (2) = f (2) - f (2) = 0

The denominator becomes

f (0 - 0^{2} + 1) - f (1) = f (1) - f (1) = 0

Since we have arrived at the indeterminate form

\frac{0}{0}

, we are justified in applying L'Hospital's Rule.

The Surgeon's Tool

L'Hospital's Rule states that for a

\frac{0}{0}

form, the limit of the ratio is equal to the limit of the ratio of the derivatives. We differentiate the numerator and denominator with respect to

h

using the Chain Rule.

For the numerator, the derivative of

f (2 h + 2 + h^{2}) - f (2)

is:

f^{'} (2 h + 2 + h^{2}) \cdot (2 + 2 h)

For the denominator, the derivative of

f (h - h^{2} + 1) - f (1)

is:

f^{'} (h - h^{2} + 1) \cdot (1 - 2 h)

Final Calculation

We now evaluate the limit of the ratio of these derivatives as

h \to 0

h \to 0 lim \frac{f ^{'} ( 2 h + 2 + h ^{2} ) \cdot ( 2 + 2 h )}{f ^{'} ( h - h ^{2} + 1 ) \cdot ( 1 - 2 h )}

Substituting

h = 0

into the expression, we obtain:

\frac{f ^{'} ( 2 ) \cdot 2}{f ^{'} ( 1 ) \cdot 1}

Using the given values

f^{'} (2) = 6

and

f^{'} (1) = 4

, the calculation yields:

\frac{6 \cdot 2}{4 \cdot 1} = \frac{12}{4} = 3

The final value of the limit is 3.

Common Traps & FAQs

Why can't I just plug in

h = 0

immediately? It gives me

0/0

, which is just 0, right?

That is the most dangerous trap in calculus!

0/0

is not zero; it is an indeterminate form. It means the limit *might* exist, but we need to do more work to uncover it. Think of it as a locked door—plugging in

h = 0

just shows you the door is locked, but L'Hospital's Rule is the key to opening it.

I always mess up the Chain Rule when there are multiple terms inside the function. How do I stay organized?

Great question. The secret is to treat the inner expression as a separate variable, say

u = 2 h + 2 + h^{2}

. Differentiate

f (u)

with respect to

u

(which is

f^{'} (u)

), and then multiply by the derivative of

u

with respect to

h

(which is

2 + 2 h

). Write them separately on the side of your paper before combining them. Never try to do it all in your head!

Do I really need L'Hospital's Rule here, or is there a definition of derivative approach?

You are absolutely right to ask! You *could* use the definition of the derivative, but it would be significantly more tedious. L'Hospital's Rule is the elegant, professional shortcut for this specific structure. In JEE Advanced, efficiency is just as important as accuracy, so embrace the shortcut when the conditions are met.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: $lim_{h \to 0} \frac{f ( 2 h + 2 + h ^{2} ) - f ( 2 )}{f ( h - h ^{2} + 1 ) - f ( 1 )}$ , given that $f^{'} (2) = 6$ and $f^{'} (1) = 4$

Select Answer:

Visualized Solution

Analyzing the Limit Form

Checking Indeterminacy $\frac{0}{0}$

Applying L'Hospital's Rule

Differentiating the Numerator

Differentiating the Denominator

Substituting $h = 0$

Using Given Values

Final Calculation

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Diagnostic

The Surgeon's Tool

Final Calculation

Similar Questions

Let $f : R \to R$ be a function such that $f (2) = 4$ and $f^{'} (2) = 1$ . Then, the value of $lim_{x \to 2} \frac{x ^{2} f ( 2 ) - 4 f ( x )}{x - 2}$ is equal to :

Let $f : R \to R$ be a differentiable function satisfying $f^{'} (3) + f^{'} (2) = 0$ . Then $lim_{x \to 0} (\frac{1 + f ( 3 + x ) - f ( 3 )}{1 + f ( 2 - x ) - f ( 2 )})^{1/ x}$ is equal to

Let $f (x) = 4$ and $f^{'} (x) = 4$ . Then $lim_{x \to 2} \frac{x f ( 2 ) - 2 f ( x )}{x - 2}$ is given by

Let $f : R \to R$ be such that $f (1) = 3$ and $f^{'} (1) = 6$ . Then $lim_{x \to 0} (\frac{f ( 1 + x )}{f ( 1 )})^{1/ x}$ equals

Let $f : R \to (0, \infty)$ be a twice differentiable function such that $f (3) = 18$ , $f^{'} (3) = 0$ and $f^{''} (3) = 4$ . Then $lim_{x \to 1} (lo g_{e} (\frac{f ( 2 + x )}{f ( 3 )}))^{\frac{18}{( x - 1 ) ^{2}}}$ is equal to :

Let $f (x)$ be a differentiable function at $x = a$ with $f^{'} (a) = 2$ and $f (a) = 4$ . Then $lim_{x \to a} \frac{x f ( a ) - a f ( x )}{x - a}$ equals :

If $f (1) = 1, f^{'} (1) = 2$ , then $lim_{x \to 1} \frac{f ( x ) - 1}{x - 1}$ is

If $f (a) = 2, f^{'} (a) = 1, g (a) = - 1, g^{'} (a) = 2$ , then the value of $lim_{x \to a} \frac{g ( x ) f ( a ) - g ( a ) f ( x )}{x - a}$ is

Let $f : R \to R$ be a continuously differentiable function such that $f (2) = 6$ and $f^{'} (2) = 1/48$ . If $\int_{6}^{f (x)} 4 t^{3} d t = (x - 2) g (x)$ , then $lim_{x \to 2} g (x)$ is equal to :

The value of $lim_{h \to 0} {\frac{3 s i n ( \frac{π}{6} + h ) - c o s ( \frac{π}{6} + h )}{3 h ( 3 c o s h - s i n h )}}$ is:

Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limh→0​f(h−h2+1)−f(1)f(2h+2+h2)−f(2)​, given that f′(2)=6 and f′(1)=4

Select Answer:

Visualized Solution

Analyzing the Limit Form

Checking Indeterminacy .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 00​

Applying L'Hospital's Rule

Differentiating the Numerator

Differentiating the Denominator

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h=0

Using Given Values

Final Calculation

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Diagnostic

The Surgeon's Tool

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a function such that f(2)=4 and f′(2)=1. Then, the value of limx→2​x−2x2f(2)−4f(x)​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a differentiable function satisfying f′(3)+f′(2)=0. Then limx→0​(1+f(2−x)−f(2)1+f(3+x)−f(3)​)1/x is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=4 and f′(x)=4. Then limx→2​x−2xf(2)−2f(x)​ is given by

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be such that f(1)=3 and f′(1)=6. Then limx→0​(f(1)f(1+x)​)1/x equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→(0,∞) be a twice differentiable function such that f(3)=18, f′(3)=0 and f′′(3)=4. Then limx→1​(loge​(f(3)f(2+x)​))(x−1)218​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a differentiable function at x=a with f′(a)=2 and f(a)=4. Then limx→a​x−axf(a)−af(x)​ equals :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(1)=1,f′(1)=2, then limx→1​x​−1f(x)​−1​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(a)=2,f′(a)=1,g(a)=−1,g′(a)=2, then the value of limx→a​x−ag(x)f(a)−g(a)f(x)​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a continuously differentiable function such that f(2)=6 and f′(2)=1/48. If ∫6f(x)​4t3dt=(x−2)g(x), then limx→2​g(x) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of limh→0​{3​h(3​cosh−sinh)3​sin(6π​+h)−cos(6π​+h)​} is:

Animated Solution for Mathematics - Limits, Continuity and Differentiability: $lim_{h \to 0} \frac{f ( 2 h + 2 + h ^{2} ) - f ( 2 )}{f ( h - h ^{2} + 1 ) - f ( 1 )}$ , given that $f^{'} (2) = 6$ and $f^{'} (1) = 4$

Checking Indeterminacy $\frac{0}{0}$

Substituting $h = 0$

Let $f : R \to R$ be a function such that $f (2) = 4$ and $f^{'} (2) = 1$ . Then, the value of $lim_{x \to 2} \frac{x ^{2} f ( 2 ) - 4 f ( x )}{x - 2}$ is equal to :

Let $f : R \to R$ be a differentiable function satisfying $f^{'} (3) + f^{'} (2) = 0$ . Then $lim_{x \to 0} (\frac{1 + f ( 3 + x ) - f ( 3 )}{1 + f ( 2 - x ) - f ( 2 )})^{1/ x}$ is equal to

Let $f (x) = 4$ and $f^{'} (x) = 4$ . Then $lim_{x \to 2} \frac{x f ( 2 ) - 2 f ( x )}{x - 2}$ is given by

Let $f : R \to R$ be such that $f (1) = 3$ and $f^{'} (1) = 6$ . Then $lim_{x \to 0} (\frac{f ( 1 + x )}{f ( 1 )})^{1/ x}$ equals

Let $f : R \to (0, \infty)$ be a twice differentiable function such that $f (3) = 18$ , $f^{'} (3) = 0$ and $f^{''} (3) = 4$ . Then $lim_{x \to 1} (lo g_{e} (\frac{f ( 2 + x )}{f ( 3 )}))^{\frac{18}{( x - 1 ) ^{2}}}$ is equal to :

Let $f (x)$ be a differentiable function at $x = a$ with $f^{'} (a) = 2$ and $f (a) = 4$ . Then $lim_{x \to a} \frac{x f ( a ) - a f ( x )}{x - a}$ equals :

If $f (1) = 1, f^{'} (1) = 2$ , then $lim_{x \to 1} \frac{f ( x ) - 1}{x - 1}$ is

If $f (a) = 2, f^{'} (a) = 1, g (a) = - 1, g^{'} (a) = 2$ , then the value of $lim_{x \to a} \frac{g ( x ) f ( a ) - g ( a ) f ( x )}{x - a}$ is

Let $f : R \to R$ be a continuously differentiable function such that $f (2) = 6$ and $f^{'} (2) = 1/48$ . If $\int_{6}^{f (x)} 4 t^{3} d t = (x - 2) g (x)$ , then $lim_{x \to 2} g (x)$ is equal to :

The value of $lim_{h \to 0} {\frac{3 s i n ( \frac{π}{6} + h ) - c o s ( \frac{π}{6} + h )}{3 h ( 3 c o s h - s i n h )}}$ is: