Let f : (-∞, ∞) - {0} → R be a differentiable function such that f'(1) = lim(a→∞) a²f(1/a). Then lim

Q: Wait, why do we reject the idea of evaluating the limit directly using L'Hopital's rule?

That is a great question! While L'Hopital's rule is a powerful tool, applying it to an expression like $\frac{a(a+1)}{2} \tan^{-1}(\frac{1}{a}) + a^2 - 2 \log_e a$ would lead to a nightmare of algebraic complexity. The problem is designed to test your ability to recognize patterns and identify the underlying function $f(x)$. By using the substitution $x = \frac{1}{a}$, we bypass the need for brute-force calculus and reveal the elegant structure of the function.

Q: How did you know that $f(x)$ was the expression inside the bracket?

We compared the given condition $f'(1) = \lim_{a \to \infty} a^2 f(\frac{1}{a})$ with the limit we were asked to evaluate. By substituting $x = \frac{1}{a}$, the limit expression became $\lim_{x \to 0^+} \frac{1}{x^2} f(x)$. By factoring out $\frac{1}{x^2}$ from our transformed expression, the remaining term inside the brackets must be $f(x)$. It is all about matching the form!

Q: Why does $\log_e a$ become $-\log_e x$?

Remember the substitution $x = \frac{1}{a}$, which means $a = \frac{1}{x}$. Using the properties of logarithms, $\log_e a = \log_e(\frac{1}{x}) = \log_e(1) - \log_e(x)$. Since $\log_e(1) = 0$, we are left with $-\log_e x$. It is a simple but crucial step that often trips students up!

Question

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

Options:

(A) $\frac{3}{2} + \frac{\pi}{4}$

(B) $\frac{3}{4} + \frac{\pi}{8}$

(C) $\frac{3}{8} + \frac{\pi}{4}$

(D) $\frac{5}{2} + \frac{\pi}{8}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are tackling a problem that looks like a monster. It is a limit at infinity involving inverse trigonometric functions, natural logarithms, and a mysterious function $f(x)$.

Many students see this and immediately think of L'Hopital's rule or Taylor series expansions. But stop! Take a breath. The JEE Advanced is not just about calculation; it is about insight.

The problem provides the condition:
$$f'(1) = \lim_{a 	o \infty} a^2 f\left(\frac{1}{a}ight)$$
This is not just a piece of information; it is the key to the entire vault. We are asked to evaluate:
$$L = \lim_{a 	o \infty} \left[ \frac{a(a+1)}{2} 	an^{-1}\left(\frac{1}{a}ight) + a^2 - 2 \log_e a ight]$$

Notice the structure? It is screaming at us that this limit is exactly the expression we need to find $f'(1)$. Our mission is not to solve the limit, but to unmask $f(x)$.

### The Power of Substitution

To simplify this, let us use the most powerful tool in our arsenal: substitution. Let $x = \frac{1}{a}$. As $a 	o \infty$, our variable $x$ gracefully approaches $0^+$.

This changes everything. The term $a^2$ becomes $\frac{1}{x^2}$, and the term $\log_e a$ becomes $\log_e\left(\frac{1}{x}ight)$, which is $-\log_e x$.

Now, let us rewrite our limit $L$ in terms of $x$:
$$L = \lim_{x 	o 0^+} \left[ \frac{\frac{1}{x}(\frac{1}{x} + 1)}{2} 	an^{-1}(x) + \frac{1}{x^2} - 2(-\log_e x) ight]$$

Simplifying this, we get:
$$L = \lim_{x 	o 0^+} \left[ \frac{1+x}{2x^2} 	an^{-1}(x) + \frac{1}{x^2} + 2 \log_e x ight]$$

### The Unmasking

Now, look at the expression again. We know $L = \lim_{x 	o 0^+} \frac{1}{x^2} f(x)$. So, let us factor out $\frac{1}{x^2}$ from our expression:
$$L = \lim_{x 	o 0^+} \frac{1}{x^2} \left[ \frac{1+x}{2} 	an^{-1}(x) + 1 + 2x^2 \log_e x ight]$$

By comparing this to the form $\lim_{x 	o 0^+} \frac{1}{x^2} f(x)$, we have successfully extracted our hidden function:
$$f(x) = \frac{1+x}{2} 	an^{-1}(x) + 1 + 2x^2 \log_e x$$

We have done the hard work! We do not need to evaluate the limit anymore. The problem tells us that this limit is equal to $f'(1)$.

### The Final Victory

Now, we just need to find the derivative of $f(x)$ and evaluate it at $x=1$. Let us differentiate:
$$f(x) = \frac{1+x}{2} 	an^{-1}(x) + 1 + 2x^2 \log_e x$$

Applying the product rule to the first term, we get:
$$\frac{1}{2} 	an^{-1}(x) + \frac{1+x}{2(1+x^2)}$$
The derivative of the constant $1$ is $0$.

For the last term, $2x^2 \log_e x$, the product rule gives us:
$$4x \log_e x + 2x^2 \cdot \frac{1}{x} = 4x \log_e x + 2x$$

Combining these, we have:
$$f'(x) = \frac{1}{2} 	an^{-1}(x) + \frac{1+x}{2(1+x^2)} + 4x \log_e x + 2x$$

Finally, we plug in $x=1$:
$$f'(1) = \frac{1}{2} 	an^{-1}(1) + \frac{1+1}{2(1+1^2)} + 4(1) \log_e(1) + 2(1)$$

Since $	an^{-1}(1) = \frac{\pi}{4}$ and $\log_e(1) = 0$, this simplifies beautifully to:
$$f'(1) = \frac{1}{2}\left(\frac{\pi}{4}ight) + \frac{2}{4} + 0 + 2 = \frac{\pi}{8} + 0.5 + 2 = \mathbf{\frac{5}{2} + \frac{\pi}{8}}$$

And there it is! The complexity vanishes, leaving behind a simple, elegant result. You have conquered the problem!

Select Answer:

Visualized Solution

Analyze the Given Condition

Change of Variable: $x = \frac{1}{a}$

Transform the Limit Expression

Define the Function $f (x)$

The Master Plan: Find $f^{'} (1)$

Differentiate $f (x)$

Differentiate the Logarithmic Term

Evaluate $f^{'} (1)$

Final Conclusion

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

Analyzing the Setup

The Power of Substitution

The Unmasking

The Final Victory

Similar Questions

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 $f : R \to R$ 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 $f (x) = lim_{n \to \infty} \sum_{r = 0}^{n} (\frac{t a n ( x / 2 ^{r + 1} ) + t a n ^{3} ( x / 2 ^{r + 1} )}{1 - t a n ^{2} ( x / 2 ^{r + 1} )}) .$ Then $lim_{x \to 0} \frac{e ^{x} - e ^{f (x)}}{x - f ( x )}$ 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 :

Let $f : [- \frac{π}{2}, \frac{π}{2}] \to R$ be a differentiable function such that $f (0) = \frac{1}{2}$ . If $lim_{x \to 0} \frac{x \int _{0}^{x} f ( t ) d t}{e ^{x^{2}} - 1} = α$ , then $8 α^{2}$ is equal to :

Find $lim_{x \to 0} {tan (π /4 + x)}^{1/ x} .$

If $lim_{x \to \infty} ((\frac{e}{1 - e}) (\frac{1}{e} - \frac{x}{1 + x}))^{x} = α$ , then the value of $\frac{l o g _{e} α}{1 + l o g _{e} α}$ equals

$lim_{x \to 0} (tan (\frac{π}{4} + x))^{\frac{1}{x}}$ is equal to :

$lim_{x \to \frac{π}{2}} \frac{1 - t a n ( x /2 )}{1 + t a n ( x /2 )} \frac{[ 1 - s i n x ]}{[ π - 2 x ] ^{3}}$ is

$lim_{x \to 0} (\frac{1 + 5 x ^{2}}{1 + 3 x ^{2}})^{1/ x^{2}} = ...$

Select Answer:

Visualized Solution

Analyze the Given Condition

Change of Variable: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=a1​

Transform the Limit Expression

Define the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The Master Plan: Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(1)

Differentiate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Differentiate the Logarithmic Term

Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(1)

Final Conclusion

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

Analyzing the Setup

The Power of Substitution

The Unmasking

The Final Victory

Similar Questions

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