Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)$ be a differentiable function at $x=a$ with $f^{\prime }(a)=2$ and $f(a)=4$. Then ${\lim }_{x\to a}\frac{xf(a)-af(x)}{x-a}$ equals :

Welcome, future engineers! Today, we are going to dissect a problem that tests not just your calculus skills, but your ability to stay calm under pressure. In the JEE Advanced arena, you will often face problems that look like a tangled mess of variables.

The secret to success is not just knowing the formulas, but understanding the story the math is trying to tell you. We are looking at a limit:

At first glance, it looks like a daunting expression. But remember, in the world of JEE Advanced, complexity is often just a mask for simplicity. Let us peel back that mask together.

Imagine a curve $y=f(x)$ plotted on your coordinate plane. We are zooming in on a specific point $x=a$.

We are given two vital pieces of information: the height of the curve at this point is $f(a)=4$, and the slope of the tangent line at this point is $f^{\prime }(a)=2$. These are our anchors.

When you see $f(a)$ in an expression, do not panic. It is just the number $4$. It is a constant. This is the most common trap students fall into—treating $f(a)$ as a variable that needs to be differentiated.

Now, let us look at the limit expression again:

The first rule of limits is always direct substitution. Let us plug in $x=a$.

The numerator becomes $a\cdot f(a)-a\cdot f(a)$, which is $0$. The denominator becomes $a-a$, which is also $0$. We have arrived at the $\frac{0}{0}$ indeterminate form.

This is the classic JEE trap, but it is also a gift. It is the universe's way of telling you that the limit exists and that you have a clear path forward: L'Hopital's Rule.

L'Hopital's Rule is a powerful tool, but it requires precision. We must differentiate the numerator and the denominator separately with respect to $x$.

Do not make the mistake of applying the quotient rule here! That would lead you into a labyrinth of unnecessary algebra. Instead, let us focus on the numerator: $xf(a)-af(x)$.

We are differentiating with respect to $x$. The first term is $x\cdot f(a)$. Since $f(a)$ is a constant (the number $4$), the derivative of $x\cdot f(a)$ is simply $f(a)$.

The second term is $-a\cdot f(x)$. Here, $a$ is a constant multiplier, so the derivative is $-a\cdot f^{\prime }(x)$.

Now, look at the denominator: $x-a$. The derivative of $x$ is $1$, and the derivative of the constant $a$ is $0$. So, the derivative of the denominator is just $1$.

Our limit has transformed from a terrifying fraction into something much more manageable:

We are almost there. Now that we have simplified the expression, we can perform the direct substitution again.

As $x$ approaches $a$, the term $f^{\prime }(x)$ approaches $f^{\prime }(a)$. So, our limit becomes:

We have all these values! We were given $f(a)=4$ and $f^{\prime }(a)=2$. Substituting these in, we get $L=4-a(2)$, which simplifies to $4-2a$.

That is it! You have conquered the limit. It was not about complex integration or series expansion; it was about understanding the definition of the derivative and the power of L'Hopital's Rule.