Let f(x) = 4 and f'(x) = 4. Then lim (x-2) (xf(2) - 2f(x)) / (x - 2) is given by (A) 2 (B) -2 (C) -

Q: Wait, why is $f(2)$ treated as a constant?

Great question! In the expression $x \cdot f(2)$, $f(2)$ is just the value of the function at the specific point $x=2$. Since it is a fixed number, it behaves like any other constant, such as $5$ or $10$. When we differentiate with respect to $x$, the derivative of a constant times $x$ is simply the constant itself.

Q: Why can't we just use energy conservation?

Energy conservation is a principle for physical systems, but here we are dealing with pure calculus! This is a limit problem, not a mechanics problem. We use the tools of calculus—specifically L'Hopital's Rule—to navigate the indeterminate form.

Q: Why did we get $-4$ instead of $0$?

It is easy to assume the answer is $0$ because the expression looks like it might vanish, but the $x$ term in the numerator changes everything. By applying L'Hopital's Rule, we find the rate of change of the numerator relative to the denominator, which leads us to the correct value of $-4$.

Question

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

Options:

(A) 2

(B) -2

(C) -4

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to unravel a limit problem that might seem intimidating at first glance, but beneath its surface lies a beautiful, logical structure.

We are given a function $f(x)$ with the conditions $f(2) = 4$ and $f'(2) = 4$. We need to evaluate the following limit:

$$\lim_{x 	o 2} \frac{x f(2) - 2 f(x)}{x - 2}$$

When you first look at this, your instinct might be to panic. But take a deep breath; in the world of limits, the first step is always to test the waters.

Let us substitute $x = 2$ directly into the expression. The numerator becomes $2 \cdot f(2) - 2 \cdot f(2)$, which is $0$. The denominator becomes $2 - 2$, which is also $0$.

We have arrived at the infamous $\frac{0}{0}$ indeterminate form. This is not a dead end; it is a gateway!

### The Lighthouse: L'Hopital's Rule

Whenever you see $\frac{0}{0}$ or $\frac{\infty}{\infty}$, think of **L'Hopital's Rule** as your lighthouse. It tells us that the limit of the ratio of two functions is equal to the limit of the ratio of their derivatives.

So, we must differentiate the numerator and the denominator separately with respect to $x$. Do not fall into the trap of using the quotient rule here, as that would be a massive waste of time.

Instead, let us differentiate the numerator $g(x) = x \cdot f(2) - 2 \cdot f(x)$ and the denominator $h(x) = x - 2$ independently.

### The Art of Differentiation

Let us focus on the numerator first. We have $x \cdot f(2) - 2 \cdot f(x)$. Remember, $f(2)$ is just a constant value, a fixed number.

When we differentiate $x \cdot f(2)$ with respect to $x$, we get $f(2)$. For the second term, the derivative of $2 \cdot f(x)$ is simply $2 \cdot f'(x)$.

So, the derivative of our numerator is $f(2) - 2 \cdot f'(x)$. Now, for the denominator, $x - 2$, the derivative is simply $1$.

It is elegant, is it not? We have transformed our terrifying limit into:

$$\lim_{x 	o 2} \frac{f(2) - 2 f'(x)}{1}$$

### The Final Calculation

Now that the indeterminate form is gone, we can safely substitute $x = 2$ into our new expression. This gives us $f(2) - 2 \cdot f'(2)$.

We were given the values at the very start: $f(2) = 4$ and $f'(2) = 4$. Substituting these in, we get:

$$4 - 2(4) = 4 - 8 = -4$$

The final result is **$-4$**. You see? By staying calm and applying the right tools, we turned a complex problem into a simple arithmetic calculation.

Keep practicing, keep questioning, and most importantly, keep falling in love with the elegance of mathematics!

Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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

Select Answer:

Visualized Solution

Analyze the Given Information

Check for Indeterminate Form

Apply L'Hopital's Rule

Differentiate the Numerator

Differentiate the Denominator

Apply the Limit to the Derivatives

Final Calculation

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

Analyzing the Setup

The Lighthouse

The Art of Differentiation

The Final Calculation

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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; } Let f(x)=4 and f′(x)=4. Then limx→2​x−2xf(2)−2f(x)​ is given by