Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f:R\to R$ be such that $f(1)=3$ and $f^{\prime }(1)=6$. Then ${\lim }_{x\to 0}{\left(\frac{f(1+x)}{f(1)}\right)}^{1/x}$ equals

Imagine you are standing at the point $x=1$ on the graph of a function $f(x)$. You know two things: the function passes through the point $(1,3)$, so $f(1)=3$, and the slope of the tangent line at that point is $6$, meaning $f^{\prime }(1)=6$.

Now, you are asked to evaluate a limit that looks quite intimidating:

At first glance, the variable $x$ in the exponent makes it look like a monster. But in the world of JEE Advanced, every monster has a weakness.

The first step is always to check the form of the limit. As $x\to 0$, the base $\frac{f(1+x)}{f(1)}$ approaches $\frac{f(1)}{f(1)}=1$, and the exponent $\frac{1}{x}$ approaches $\infty$. We have encountered the classic $1^{\infty }$ indeterminate form.

Whenever you see a limit of the form ${\lim }_{x\to a}[f(x){]}^{g(x)}$ resulting in $1^{\infty }$, you have a secret weapon. The formula is:

This formula is a bridge that turns a terrifying exponentiation problem into a simple multiplication problem. Let us apply this to our limit.

Here, our base is $\frac{f(1+x)}{f(1)}$ and our exponent is $\frac{1}{x}$. Plugging these into our formula, we get:

Now, the problem becomes an exercise in algebraic manipulation. We need to simplify the expression inside the bracket.

By finding a common denominator, we get:

Substituting this back into our limit, we have:

Since $f(1)$ is just a constant value of $3$, we can pull $\frac{1}{f(1)}$ outside the limit:

Look closely at the limit that remains: ${\lim }_{x\to 0}\frac{f(1+x)-f(1)}{x}$. Does this look familiar? It is the fundamental definition of the derivative $f^{\prime }(1)$ from first principles!

We were given $f^{\prime }(1)=6$ right at the start. The entire limit has collapsed into a simple ratio. We now have:

Substituting our known values, $f^{\prime }(1)=6$ and $f(1)=3$, we get:

It is truly elegant how the complexity of the limit dissolves into the simple ratio of the derivative to the function value. Keep this technique in your toolkit; whenever you see $1^{\infty }$, remember the magic formula and look for the hidden derivative! The final answer is $e^2$.