Animated Solution for Mathematics - Limits, Continuity and Differentiability: If ${\lim }_{x\to 0}(\frac{\tan x}{x}{)}^{\frac{1}{x^2}}=p$, then $96{\log }_{e}p$ is equal to ______

Welcome, fellow traveler on the JEE journey! Today, we are going to demystify a problem that often intimidates students: the limit of a function raised to another function, specifically the $1^{\infty }$ indeterminate form.

Imagine you are standing before the expression $p={\lim }_{x\to 0}(\frac{\tan x}{x}{)}^{\frac{1}{x^2}}$. At first glance, it looks like a mountain. But as we peel back the layers, you will see it is just a beautiful, elegant dance of algebra.

When we see a limit of the form $1^{\infty }$, our first instinct should be to use the powerful exponential identity:

This is our secret weapon. It transforms the terrifying exponential structure into a much more manageable limit in the exponent.

Here, our base $f(x)$ is $\frac{\tan x}{x}$ and our exponent $g(x)$ is $\frac{1}{x^2}$. Substituting these into our identity, we get:

Now, look at the exponent: $\frac{1}{x^2}(\frac{\tan x}{x}-1)$. If we simplify this, we get:

This is where many students reach for L'Hopital's rule, differentiating repeatedly until they get lost in a sea of trigonometric derivatives. But we are smarter than that!

We will use the Taylor series expansion for $\tan x$ near $x=0$, which is:

This expansion acts like a microscope, allowing us to see the behavior of the function right at the point of interest.

Let's substitute this series into our expression:

Notice the magic? The $x$ terms cancel out perfectly, leaving us with:

When we divide by $x^3$, we are left with $\frac{1}{3}+\text{higher order terms}$. As $x$ approaches zero, all those higher order terms vanish into thin air, leaving us with exactly $\frac{1}{3}$.

So, our limit in the exponent is $\frac{1}{3}$, which means $p=e^{1/3}$. The question asks for $96{\log }_{e}p$.

Since ${\log }_{e}p={\log }_e(e^{1/3})=\frac{1}{3}$, the final calculation is simply:

There you have it! We navigated the $1^{\infty }$ form, used the Taylor series to simplify the expression, and arrived at the answer with grace. Keep practicing, and remember: every complex problem is just a series of simple steps waiting to be discovered. The final answer is 32.