If f(a) = 2, f'(a) = 1, g(a) = -1, g'(a) = 2, then the value of lim (x→a) (g(x)f(a) - g(a)f(x))/(x -

Q: Wait, why do we treat $f(a)$ as a constant? Isn't it a function?

That is a brilliant question! It is easy to get confused because $f$ is a function. However, look closely at the notation: $f(a)$ is the value of the function $f$ evaluated at the specific point $a$. Since $a$ is a fixed constant, $f(a)$ is just a number. When we differentiate with respect to $x$, any constant multiplier simply stays put, just like the $5$ in $\frac{d}{dx}(5x^2)$.

Q: Why can't we just use the product rule on $g(x)f(a)$?

You absolutely could, but you would find that the derivative of $f(a)$ with respect to $x$ is zero! If you apply the product rule, you get $g'(x)f(a) + g(x)f'(a)$. But since $f(a)$ is a constant, $f'(a)$ is not a derivative with respect to $x$; it is just a value. The derivative of a constant is zero. So, $g'(x)f(a) + g(x) \cdot 0$ leaves you with exactly what we derived: $g'(x)f(a)$.

Q: Is L'Hopital's Rule always the best way to solve these?

In the context of JEE Advanced, L'Hopital's Rule is a fantastic, reliable tool for $0/0$ or $\infty/\infty$ forms. While you could also solve this by adding and subtracting $g(a)f(a)$ in the numerator to force the definition of the derivative, L'Hopital's Rule is often faster and less prone to algebraic errors. Master both, but use the one that feels most intuitive to you in the exam hall!

Question

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

Options:

(A) $-5$

(B) $1/5$

(C) 5

(D) none of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to dissect a problem that looks like a tangled mess of variables but is actually a beautiful, elegant dance of calculus.

We are tasked with evaluating the limit:

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

Given the values $f(a) = 2$, $f'(a) = 1$, $g(a) = -1$, and $g'(a) = 2$. At first glance, this might look intimidating, but in the world of JEE Advanced, the most complex-looking problems are often the ones that yield the most satisfying solutions once you find the right key.

### The Diagnostic

Before we start throwing complex theorems at the problem, we must perform a diagnostic. In calculus, whenever we see a limit, our first instinct must always be direct substitution.

Let us see what happens when we plug $x = a$ into our expression. The numerator becomes $g(a)f(a) - g(a)f(a)$, which is clearly $0$. The denominator becomes $a - a$, which is also $0$.

We have arrived at the classic $0/0$ indeterminate form. This is not a dead end; it is a green light! It tells us that the limit exists and that we have a powerful tool at our disposal: **L'Hopital's Rule**.

### The Power of L'Hopital

L'Hopital's Rule is our scalpel. It allows us to resolve the $0/0$ ambiguity by differentiating the numerator and the denominator separately with respect to $x$.

Here is where many students stumble, so let us be precise. We are differentiating with respect to $x$, which means $f(a)$ and $g(a)$ are not variables; they are fixed constants.

When we differentiate the numerator $g(x)f(a) - g(a)f(x)$, we treat $f(a)$ and $g(a)$ as coefficients. Thus, the derivative of the numerator becomes:

$$ \frac{d}{dx}[g(x)f(a) - g(a)f(x)] = g'(x)f(a) - g(a)f'(x) $$

Now, look at the denominator. The derivative of $x - a$ with respect to $x$ is simply $1 - 0 = 1$. The complexity has vanished!

### The Final Calculation

With the indeterminate form resolved, our limit now simplifies to:

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

Now, we can safely substitute $x = a$ into this expression. We get:

$$ g'(a)f(a) - g(a)f'(a) $$

We are given all these values in the problem statement. Let us plug them in with care: $f(a) = 2$, $f'(a) = 1$, $g(a) = -1$, and $g'(a) = 2$.

Substituting these, we get:

$$ (2)(2) - (-1)(1) $$

This simplifies to $4 - (-1)$, which is $4 + 1 = 5$.

### The Takeaway

Look at that! We started with a terrifying-looking limit and ended with a simple integer.

The secret was not in complex manipulation, but in recognizing the indeterminate form and correctly identifying which terms were constants and which were functions. Keep this clarity of mind, and you will conquer any problem the JEE throws at you.

The final result is **5**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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

Select Answer:

Visualized Solution

Given Values

Checking Limit Form: Numerator

Checking Limit Form: Denominator

Indeterminate Form $\frac{0}{0}$

Applying L'Hopital's Rule

Derivative of Numerator

Derivative of Denominator

Limit After L'Hopital

Substituting $x = a$

Substituting Given Values

Final Answer

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

Analyzing the Setup

The Diagnostic

The Power of L'Hopital

The Final Calculation

The Takeaway

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