JEE Main 2003

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f (a) = g (a) = k$ and their $n$ th derivatives $f^{n} (a), g^{n} (a)$ exist and are not equal for some $n$ . Further if $lim_{x \to a} \frac{f ( a ) g ( x ) - f ( a ) - g ( a ) f ( x ) + f ( a )}{g ( x ) - f ( x )} = 4$ then the value of $k$ is

Q: Wait, what if $g'(a) = f'(a)$? Does the method fail?

That is a brilliant observation! If $g'(a) = f'(a)$, the expression remains $\frac{0}{0}$ even after the first differentiation. That is exactly why the problem provides the condition about the $n$th derivative. It guarantees that there exists some order of differentiation where the derivatives are not equal, allowing the limit to resolve. You just keep applying L'Hopital's Rule until the terms no longer vanish!

Q: Why is $f(a)$ treated as a constant? Isn't $f$ a function?

Ah, the classic trap! $f(a)$ is the *value* of the function at a specific point $a$. It is a fixed number, $k$. When you differentiate with respect to $x$, the derivative of a constant is zero. Do not confuse $f(x)$ (the variable function) with $f(a)$ (the constant value). Keep your focus on the variable $x$!

Select Answer:

Visualized Solution

Given Conditions

Given: $f (a) = g (a) = k$
Given: $f^{(n)} (a) \neq = g^{(n)} (a)$ for some $n$

The Limit Expression

Limit expression: $L = lim_{x \to a} \frac{f ( a ) g ( x ) - f ( a ) - g ( a ) f ( x ) + f ( a )}{g ( x ) - f ( x )} = 4$

Simplifying the Numerator

Notice the terms $- f (a)$ and $+ f (a)$ in the numerator.
They cancel each other out.
Simplified limit: $L = lim_{x \to a} \frac{f ( a ) g ( x ) - g ( a ) f ( x )}{g ( x ) - f ( x )}$

Checking the Indeterminate Form

Substitute $x = a$ to check the form.
Numerator: $f (a) g (a) - g (a) f (a) = k^{2} - k^{2} = 0$
Denominator: $g (a) - f (a) = k - k = 0$
Form is $\frac{0}{0}$

Applying L'Hopital's Rule

Since the form is $\frac{0}{0}$ , we use L'Hopital's Rule.
Differentiate the numerator and denominator separately with respect to $x$ .

Differentiating the Numerator

$\frac{d}{d x} [f (a) g (x) - g (a) f (x)]$
$f (a)$ and $g (a)$ are constants.
Derivative is $f (a) g^{'} (x) - g (a) f^{'} (x)$

Differentiating the Denominator

$\frac{d}{d x} [g (x) - f (x)]$
Derivative is $g^{'} (x) - f^{'} (x)$

The New Limit Expression

$L = lim_{x \to a} \frac{f ( a ) g ^{'} ( x ) - g ( a ) f ^{'} ( x )}{g ^{'} ( x ) - f ^{'} ( x )}$

Substituting the Constants

We know $f (a) = k$ and $g (a) = k$ .
Substitute these into the limit.
$L = lim_{x \to a} \frac{k \cdot g ^{'} ( x ) - k \cdot f ^{'} ( x )}{g ^{'} ( x ) - f ^{'} ( x )}$

Factoring Out $k$

Factor $k$ from the numerator:
$L = lim_{x \to a} \frac{k ( g ^{'} ( x ) - f ^{'} ( x ))}{g ^{'} ( x ) - f ^{'} ( x )}$

The Role of the $n$ -th Derivative

If $g^{'} (a) \neq = f^{'} (a)$ , the terms cancel directly.
If $g^{'} (a) = f^{'} (a)$ , we apply L'Hopital's Rule repeatedly.
Since $f^{(n)} (a) \neq = g^{(n)} (a)$ , eventually the terms will not be zero and will cancel out.

Evaluating the Limit

The common terms cancel out: $\frac{g ^{(n)} ( x ) - f ^{(n)} ( x )}{g ^{(n)} ( x ) - f ^{(n)} ( x )} = 1$
Therefore, the limit evaluates to $k$ .
$L = k$

Final Conclusion

We are given that $L = 4$ .
We found that $L = k$ .
Therefore, $k = 4$ .

00:00 / 00:00

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

Analyzing the Setup

We are tasked with evaluating the limit:

x \to a lim \frac{f ( a ) g ( x ) - f ( a ) - g ( a ) f ( x ) + f ( a )}{g ( x ) - f ( x )} = 4

At first glance, the expression appears daunting. However, notice that the terms

- f (a)

and

+ f (a)

in the numerator cancel each other out perfectly.

This simplifies the numerator to

f (a) g (x) - g (a) f (x)

. The expression now reads:

x \to a lim \frac{f ( a ) g ( x ) - g ( a ) f ( x )}{g ( x ) - f ( x )} = 4

The Indeterminate Form

To understand the behavior of this limit as

x \to a

, we substitute

x = a

. Given that

f (a) = k

and

g (a) = k

, the numerator becomes:

f (a) g (a) - g (a) f (a) = k \cdot k - k \cdot k = 0

Similarly, the denominator becomes

g (a) - f (a) = k - k = 0

. We have arrived at the classic

\frac{0}{0}

indeterminate form, which signals that we should apply L'Hopital's Rule.

Applying L'Hopital's Rule

Since

f (a)

and

g (a)

are constants (specifically

k

), they remain unchanged during differentiation with respect to

x

. Differentiating the numerator and denominator separately with respect to

x

, we obtain:

x \to a lim \frac{f ( a ) g ^{'} ( x ) - g ( a ) f ^{'} ( x )}{g ^{'} ( x ) - f ^{'} ( x )} = 4

Substituting the known values

f (a) = k

and

g (a) = k

into this derivative expression, we get:

x \to a lim \frac{k \cdot g ^{'} ( x ) - k \cdot f ^{'} ( x )}{g ^{'} ( x ) - f ^{'} ( x )} = 4

Final Calculation

We can now factor the constant

k

out of the numerator:

k \cdot x \to a lim \frac{g ^{'} ( x ) - f ^{'} ( x )}{g ^{'} ( x ) - f ^{'} ( x )} = 4

Provided that $g'(x) eq f'(x)$ (or generally that the

n

th derivatives are not equal), the ratio simplifies to

1

. This leaves us with the simple equation:

k \cdot 1 = 4

Thus, the value of the constant is $k = 4$ .

Common Traps & FAQs

Wait, what if

g^{'} (a) = f^{'} (a)

? Does the method fail?

That is a brilliant observation! If

g^{'} (a) = f^{'} (a)

, the expression remains

\frac{0}{0}

even after the first differentiation. That is exactly why the problem provides the condition about the

n

th derivative. It guarantees that there exists some order of differentiation where the derivatives are not equal, allowing the limit to resolve. You just keep applying L'Hopital's Rule until the terms no longer vanish!

Why is

f (a)

treated as a constant? Isn't

f

a function?

Ah, the classic trap!

f (a)

is the *value* of the function at a specific point

a

. It is a fixed number,

k

. When you differentiate with respect to

x

, the derivative of a constant is zero. Do not confuse

f (x)

(the variable function) with

f (a)

(the constant value). Keep your focus on the variable

x

Can I use Taylor Series expansion instead of L'Hopital's?

Absolutely! Taylor series is a powerful, elegant alternative. You could expand

f (x)

and

g (x)

around

a

f (x) = f (a) + f^{'} (a) (x - a) + \dots

and similarly for

g (x)

. It is a more advanced path, but it leads to the same beautiful result. L'Hopital is just the most direct route for this specific problem.

Select Answer:

Visualized Solution

Given Conditions

The Limit Expression

Simplifying the Numerator

Checking the Indeterminate Form

Applying L'Hopital's Rule

Differentiating the Numerator

Differentiating the Denominator

The New Limit Expression

Substituting the Constants

Factoring Out $k$

The Role of the $n$ -th Derivative

Evaluating the Limit

Final Conclusion

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

Analyzing the Setup

The Indeterminate Form

Applying L'Hopital's Rule

Final Calculation

Similar Questions

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

If $lim_{x \to 0} {(1 - cos \frac{x ^{2}}{2} - cos \frac{x ^{2}}{4} + cos \frac{x ^{2}}{2} cos \frac{x ^{2}}{4})} = 2^{k}$ , then the value of $k$ is

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 :

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 $g (x) = \frac{( x - 1 ) ^{n}}{l o g c o s ^{m} ( x - 1 )}; 0 < x < 2, m$ and $n$ are integers, $m \neq = 0, n > 0$ , and let $p$ be the left hand derivative of $∣ x - 1∣$ at $x = 1$ . If $lim_{x \to 1^{+}} g (x) = p$ , then

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

If the function $f (x) = \frac{c o s ( s i n x ) - c o s x}{x ^{4}}$ is continuous at each point in its domain and $f (0) = \frac{1}{k}$ , then $k$ is ____

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

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 :

Select Answer:

Visualized Solution

Given Conditions

The Limit Expression

Simplifying the Numerator

Checking the Indeterminate Form

Applying L'Hopital's Rule

Differentiating the Numerator

Differentiating the Denominator

The New Limit Expression

Substituting the Constants

Factoring Out .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

The Role of the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n-th Derivative

Evaluating the Limit

Final Conclusion

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

Analyzing the Setup

The Indeterminate Form

Applying L'Hopital's Rule

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:(−∞,∞)−{0}→R be a differentiable function such that f′(1)=lima→∞​a2f(a1​). Then lima→∞​2a(a+1)​tan−1(a1​)+a2−2loge​a is equal to

.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 :

Factoring Out $k$

The Role of the $n$ -th Derivative

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

If $lim_{x \to 0} {(1 - cos \frac{x ^{2}}{2} - cos \frac{x ^{2}}{4} + cos \frac{x ^{2}}{2} cos \frac{x ^{2}}{4})} = 2^{k}$ , then the value of $k$ is

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 :

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 $g (x) = \frac{( x - 1 ) ^{n}}{l o g c o s ^{m} ( x - 1 )}; 0 < x < 2, m$ and $n$ are integers, $m \neq = 0, n > 0$ , and let $p$ be the left hand derivative of $∣ x - 1∣$ at $x = 1$ . If $lim_{x \to 1^{+}} g (x) = p$ , then

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

If the function $f (x) = \frac{c o s ( s i n x ) - c o s x}{x ^{4}}$ is continuous at each point in its domain and $f (0) = \frac{1}{k}$ , then $k$ is ____

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

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 :