Animated Solution for Mathematics - Limits, Continuity and Differentiability: If ${\lim }_{x\to 0}\frac{\log (3+x)-\log (3-x)}{x}=k$, the value of $k$ is

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a limit problem that might look simple at first glance, but it holds a beautiful lesson in the mechanics of calculus.

We are tasked with finding the value of $k$ in the expression:

This is not just an algebraic exercise; it is a fundamental encounter with the concept of the derivative.

In the world of limits, our first instinct should always be direct substitution. It is the simplest, most honest way to see if a function behaves well at a point.

Let us plug $x=0$ into our expression. The numerator becomes $\log (3+0)-\log (3-0)$, which simplifies to $\log 3-\log 3$, or $0$.

The denominator is simply $x$, which becomes $0$. We have hit the infamous $\frac{0}{0}$ indeterminate form. This is not a dead end; it is a signal that the limit exists and is waiting to be uncovered.

When we face $\frac{0}{0}$, we call upon the legendary L'Hopital's Rule. It is a powerful tool that allows us to find the limit of a quotient by differentiating the numerator and the denominator separately.

We need to compute the derivative of the numerator:

And the derivative of the denominator:

This is where the battle is won or lost. Let us differentiate the numerator term by term.

For the first term, $\log (3+x)$, the derivative is $\frac{1}{3+x}$ multiplied by the derivative of the inside, which is $1$. So, we get $\frac{1}{3+x}$.

Now, for the second term, $\log (3-x)$. The derivative of $\log (u)$ is $\frac{1}{u}$, but we must apply the Chain Rule. The derivative of the inner function $(3-x)$ is $-1$.

Therefore, the derivative of $\log (3-x)$ is:

Now, let us put it all together. Our limit expression becomes:

Notice the double negative? Subtracting a negative is the same as adding a positive. The expression simplifies beautifully to:

The denominator of $1$ has vanished, and the indeterminate form has been defeated. We are left with a clean, well-behaved expression.

With the indeterminate form gone, we can safely perform direct substitution once more. Plugging $x=0$ into our simplified expression, we get:

The final result is $k=\frac{2}{3}$.

It is elegant, it is precise, and it is the result of careful, step-by-step reasoning. Remember, the beauty of calculus lies not just in the final answer, but in the journey of transforming a complex, undefined expression into a simple, concrete value.