Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to \frac{\pi }{2}}\left({\tan }^{2}x\left((2{\sin }^{2}x+3\sin x+4{)}^{\frac{1}{2}}-({\sin }^{2}x+6\sin x+2{)}^{\frac{1}{2}}\right)\right)$ is equal to

Taming the Indeterminate

Welcome, fellow traveler on the path to JEE mastery. Today, we are standing before a classic, intimidating limit problem.

It looks like a monster, doesn't it? We have a product of a trigonometric function shooting off to infinity and a difference of two square roots that seem to vanish into zero.

This is the classic $\infty \times 0$ indeterminate form. But fear not—in the world of JEE Advanced, every monster has a weakness. Let us break this down, step by step, and turn this complexity into a beautiful, simple fraction.

The Indeterminate Trap

Our first step is always to verify the form. As $x\to \frac{\pi }{2}$, we know that $\sin x\to 1$ and $\tan x\to \infty$.

If we look inside the parentheses, we have:

We are indeed staring at an $\infty \times 0$ form. We cannot evaluate this directly; we need to perform some algebraic surgery.

The Conjugate Weapon

When you see square roots in a limit, the conjugate is your best friend. We multiply and divide the entire expression by the conjugate:

By doing this, we invoke the difference of squares identity: $(a-b)(a+b)=a^2-b^2$.

This will strip away those radical signs in the numerator, leaving us with a much friendlier polynomial expression. The denominator, meanwhile, will just be the sum of the two square roots, which we already know evaluates to $3+3=6$ as $x\to \frac{\pi }{2}$.

Algebraic Surgery

Now, let us focus on the numerator. After the rationalization, we have:

Distributing that negative sign is crucial—don't let a simple sign error ruin your hard work! Simplifying this, we get ${\sin }^{2}x-3\sin x+2$.

This is a quadratic in terms of $\sin x$. Factoring it is straightforward: we look for two numbers that multiply to $2$ and add to $-3$. Those are $-1$ and $-2$. Thus, our numerator becomes $(\sin x-1)(\sin x-2)$.

The Trigonometric Bridge

We are almost there. We have ${\tan }^{2}x$ multiplied by our factored numerator, all divided by $6$. As $x\to \frac{\pi }{2}$, the term $(\sin x-2)$ approaches $-1$.

We can pull this constant out. Now we are left with ${\tan }^{2}x(1-\sin x)$.

We still have an indeterminate form because ${\tan }^{2}x\to \infty$ and $(1-\sin x)\to 0$. To break this, we convert ${\tan }^{2}x$ into $\frac{{\sin }^{2}x}{{\cos }^{2}x}$. Then, we use the Pythagorean identity ${\cos }^{2}x=1-{\sin }^{2}x$.

By writing $1-{\sin }^{2}x$ as $(1-\sin x)(1+\sin x)$, we see the term $(1-\sin x)$ in both the numerator and the denominator. We cancel it out, and the indeterminate form vanishes!

We are left with:

Substituting $x=\frac{\pi }{2}$, we get:

We have conquered the beast. Remember, in JEE, it is never about memorizing the answer; it is about mastering the process of simplification. Keep practicing, and keep that curiosity alive!