Animated Solution for Mathematics - Limits, Continuity and Differentiability: Given below are two statements : Statement I : ${\lim }_{x\to 0}(\frac{{\tan }^{-1}x+{\log }_e\sqrt{\frac{1+x}{1-x}}-2x}{x^5})=\frac{2}{5}$. Statement II : ${\lim }_{x\to 1}(x^{\frac{2}{1-x}})=\frac{1}{e^2}$. In the light of the above statements, choose the correct answer from the options given below :

A Journey Through Taylor Series and Indeterminate Forms

Welcome, future engineers! Today, we are going to dismantle two intimidating limits that often appear on the JEE Advanced paper.

Many students see these problems and immediately panic, reaching for the brute force of L'Hopital's Rule. But today, we are going to learn the difference between 'solving' a problem and 'mastering' it. We will use the elegance of Taylor Series and the precision of standard limit formulas to prove these statements.

The Power of Taylor Series

Let us look at Statement I:

When you see a denominator like $x^5$, your brain should immediately scream 'Taylor Series!' If you try to differentiate this numerator five times, you will likely lose your way in a sea of chain rules and product rules.

Instead, let us expand each term individually. First, the inverse tangent:

This is a beautiful, alternating series of odd powers. We stop at $x^5$ because our denominator is $x^5$; anything higher will simply divide to zero.

Next, consider the logarithmic term: $\ln \sqrt{\frac{1+x}{1-x}}$. Using the properties of logarithms, we can rewrite this as $\frac{1}{2}[\ln (1+x)-\ln (1-x)]$.

Now, we use the standard expansions for $\ln (1+x)$ and $\ln (1-x)$:

When we subtract these, the even powers cancel out, and the odd powers double! We get $2(x+\frac{x^3}{3}+\frac{x^5}{5})$. Multiplying by the $\frac{1}{2}$ outside, we are left with $x+\frac{x^3}{3}+\frac{x^5}{5}$.

Now, combine everything in the numerator:

The $x$ terms cancel ($x+x-2x=0$), the $x^3$ terms cancel ($-\frac{x^3}{3}+\frac{x^3}{3}=0$), and we are left with $\frac{x^5}{5}+\frac{x^5}{5}=\frac{2x^5}{5}$. Dividing by $x^5$, we get exactly $\frac{2}{5}$. Statement I is true!

The $1^{\infty }$ Indeterminate Form

Now, let us tackle Statement II:

When $x\to 1$, the base $x$ approaches $1$, and the exponent $\frac{2}{1-x}$ approaches $\infty$. This is the classic $1^{\infty }$ form.

We have a powerful shortcut for this:

Here, $f(x)=x$ and $g(x)=\frac{2}{1-x}$. Applying the formula, the exponent becomes:

Notice that $(x-1)$ and $(1-x)$ are negatives of each other. Thus, $\frac{x-1}{1-x}=-1$.

The limit of the exponent is simply $2\times (-1)=-2$. Therefore, the entire limit evaluates to $e^{-2}$, which is $\frac{1}{e^2}$. Statement II is also true!

Both statements are true. By choosing the right tool—Taylor Series for polynomials and log expansions, and the standard formula for indeterminate powers—we turned a complex problem into a simple, logical flow. Keep practicing these expansions; they are the secret weapons of top JEE rankers!