Let f : R → R be a differentiable function satisfying f'(3) + f'(2) = 0. Then lim(x→0) ((1+f(3+x)-f(

Q: Wait, why do we reject the negative root here? Or rather, why is the limit $1^{\infty}$?

Great question! We aren't rejecting a root; we are identifying the indeterminate form. When you plug $x=0$ into the base $\frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)}$, you get $\frac{1+f(3)-f(3)}{1+f(2)-f(2)} = \frac{1}{1} = 1$. Since the exponent is $1/x$, as $x \to 0$, the exponent approaches $\infty$. Thus, it is the classic $1^{\infty}$ form.

Q: Why can't we just use L'Hopital's Rule directly on the original expression?

You *could* try, but it would be a nightmare of chain rules and quotient rules. The $1^{\infty}$ formula $e^{\lim h(x)[g(x)-1]}$ is a shortcut specifically designed to bypass the complexity of differentiating the entire expression. Always look for the structure first!

Q: How do I handle the negative sign in $f(2-x)$? Does it change the derivative definition?

It does! Remember that $f'(2) = \lim_{h \to 0} \frac{f(2+h)-f(2)}{h}$. Here, we have $\frac{f(2-x)-f(2)}{x}$. If we let $h = -x$, then as $x \to 0$, $h \to 0$. The expression becomes $\frac{f(2+h)-f(2)}{-h}$, which is exactly $-f'(2)$. Just be careful with that negative sign!

Question

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

Options:

(A) $e^2$

(B) $e$

(C) $e^{-1}$

(D) $1$

Sigma Percentile · Accepted Answer

### The Wolf in Sheep's Clothing: Unmasking the Limit

Imagine you are standing before a massive, intimidating mountain. At first glance, the expression

$$\lim_{x 	o 0} \left( \frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)} ight)^{1/x}$$

looks like a jagged peak of complex function notation. But in the world of JEE Advanced, we know that the most terrifying-looking problems often hide the most elegant, simple truths.

Let's break this down together.

### Phase 1: The Indeterminate Form

Before we start climbing, we must check the terrain. Whenever you see a limit with a variable in the exponent, your internal alarm should ring: is this a $1^{\infty}$ form?

As $x 	o 0$, the base becomes

$$\frac{1+f(3)-f(3)}{1+f(2)-f(2)} = \frac{1}{1} = 1$$

The exponent $1/x$ shoots off to infinity. We have confirmed it: this is a $1^{\infty}$ indeterminate form.

### Phase 2: The Master Key

We don't need to reinvent the wheel. We have a powerful tool for this: the standard limit result for $1^{\infty}$.

If we have $\lim_{x 	o a} [g(x)]^{h(x)}$, it evaluates to $e^{\lim_{x 	o a} h(x)[g(x)-1]}$. Here, our $g(x)$ is that bulky fraction, and $h(x)$ is $1/x$.

Let's set it up:

$$L = e^{\lim_{x 	o 0} \frac{1}{x} \left[ \frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)} - 1 ight]}$$

I know, it looks messy. But take a breath. The beauty of calculus is that it rewards patience.

### Phase 3: Algebraic Surgery

We need a common denominator. When we subtract $1$, the expression becomes:

$$\frac{1+f(3+x)-f(3) - [1+f(2-x)-f(2)]}{1+f(2-x)-f(2)}$$

Watch the magic happen. The $1$ and $-1$ cancel out instantly. We are left with a numerator of $[f(3+x)-f(3)] - [f(2-x)-f(2)]$.

This is the moment where the problem stops being algebra and starts being calculus. We are essentially looking at the difference of two functions.

### Phase 4: The Calculus Insight

We can split our limit into two parts. The denominator part, $\frac{1}{1+f(2-x)-f(2)}$, simply approaches $1$ as $x 	o 0$ because $f$ is continuous.

Now, look at the numerator terms:

$$\frac{f(3+x)-f(3)}{x} 	o f'(3)$$

The second term, $\frac{f(2-x)-f(2)}{x}$, requires a tiny bit of care. By substituting $h = -x$, we transform it into $-f'(2)$.

Putting it all together, our exponent becomes $f'(3) - (-f'(2))$, which is $f'(3) + f'(2)$.

### The Grand Finale

Look back at the problem statement. We were given the golden key: $f'(3) + f'(2) = 0$.

Our entire, terrifying exponent collapses into a beautiful, silent zero. And what is $e^0$?

It is $1$. The mountain has been climbed, and the view from the top is perfectly clear. The answer is **$1$**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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

Select Answer:

Visualized Solution

Identify the Limit Form

Apply the $1^{\infty}$ Formula

Apply the Formula

Simplify the Expression

Rearrange for Derivatives

Evaluate the Denominator Limit

Evaluate the Derivative Limits

Combine and Apply Given Condition

Final Conclusion

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

The Wolf in Sheep's Clothing

Phase 1

Phase 2

Phase 3

Phase 4

The Grand Finale

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