Animated Solution for Mathematics - Limits, Continuity and Differentiability: If the function $f(x)=\frac{e^x(e^{\tan x-x}-1)+{\log }_e(\sec x+\tan x)-x}{\tan x-x}$ is continuous at $x=0$, then the value of $f(0)$ is equal to

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a problem that, at first glance, looks like a tangled mess of exponentials and logarithms.

But remember, in the world of calculus, complexity is often just a mask for elegance. Our goal is to find the value of $f(0)$ for a function that is continuous at $x=0$.

The very definition of continuity is our North Star: for a function to be continuous at a point, the value of the function at that point must be equal to the limit of the function as it approaches that point. So, our mission is simple: calculate $f(0)={\lim }_{x\to 0}f(x)$.

Before we rush into the heavy machinery of calculus, let's simplify the terrain. The numerator is $e^x(e^{\tan x-x}-1)+\ln (\sec x+\tan x)-x$.

If we distribute that $e^x$, we get $e^{x+\tan x-x}-e^x$, which simplifies beautifully to $e^{\tan x}-e^x$. Now our limit looks like this:

If you try to plug in $x=0$ now, you get $\frac{0}{0}$. This is the classic indeterminate form. It is not a dead end; it is an invitation to use L'Hopital's Rule.

L'Hopital's Rule tells us that for a $\frac{0}{0}$ form, we can differentiate the numerator and the denominator separately. Let's tackle the denominator first:

Now for the numerator: the derivative of $e^{\tan x}$ is $e^{\tan x}{\sec }^{2}x$, the derivative of $-e^x$ is $-e^x$, the derivative of $\ln (\sec x+\tan x)$ is $\sec x$, and the derivative of $-x$ is $-1$.

Putting it all together, our new numerator is $e^{\tan x}{\sec }^{2}x-e^x+\sec x-1$. Let's use a clever trick by substituting ${\sec }^{2}x=1+{\tan }^{2}x$:

Now, we split the limit into three parts, all over ${\tan }^{2}x$. The first part is:

The second part is ${\lim }_{x\to 0}\frac{e^{\tan x}-e^x}{{\tan }^{2}x}$. This term vanishes to zero because the numerator is of a higher order than the denominator.

Finally, the third part is ${\lim }_{x\to 0}\frac{\sec x-1}{{\tan }^{2}x}$. Converting to sine and cosine, we get:

Since $\frac{1-\cos x}{{\sin }^{2}x}=\frac{1}{1+\cos x}$, the limit becomes $\frac{1}{2}\cdot 1=\frac{1}{2}$.

We have arrived at the finish line. Summing our three parts, we get:

The complexity has vanished, leaving behind a clean, elegant result. Remember, in JEE Advanced, the key is not just to calculate, but to simplify and strategize.

You have successfully navigated the limit, and the value of $f(0)$ is $\frac{3}{2}$. Keep practicing this mindset, and no problem will ever be too daunting.