Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to \infty }\frac{(\sqrt{3x+1}+\sqrt{3x-1}{)}^6+(\sqrt{3x+1}-\sqrt{3x-1}{)}^6}{(x+\sqrt{x^2-1}{)}^6+(x-\sqrt{x^2-1}{)}^6}{x}^3$

Imagine you are standing before a massive, intimidating algebraic expression. It is a limit problem, and $x$ is racing toward infinity.

Your first instinct might be to panic, to start expanding those binomials raised to the power of six. But stop. Take a breath.

In the world of JEE Advanced, the most complex-looking problems are often the ones that hide the most elegant, simple solutions. This problem is a classic example of algebraic camouflage. The key is not to fight the monster, but to understand its anatomy.

We are looking at a ratio of two expressions, both raised to the sixth power, multiplied by $x^3$. The secret to limits at infinity is identifying the dominant term.

As $x$ grows without bound, the smaller terms—like the $1$ in $3x+1$ or the $-1$ in $x^2-1$—become insignificant. They are like pebbles in the ocean of infinity. We need to perform a surgical extraction.

Look at the numerator: $(\sqrt{3x+1}+\sqrt{3x-1}{)}^6$. Inside the radical, we have $3x$. If we factor out $x$, we get $\sqrt{x(3+\frac{1}{x})}$.

When we pull that $\sqrt{x}$ out of the radical, it carries the power of $6$ with it. Since $(\sqrt{x}{)}^6=x^3$, we have successfully extracted an $x^3$ from the numerator.

Now, look at the denominator: $(x+\sqrt{x^2-1}{)}^6$. Inside the radical, the dominant term is $x^2$. Factoring it out gives us $x\sqrt{1-\frac{1}{x^2}}$.

When we factor $x$ out of the entire bracket $(x+x\sqrt{1-\frac{1}{x^2}}{)}^6$, it comes out as $x^6$.

This is where the magic happens. We have an $x^3$ from our numerator extraction, and there was already an $x^3$ waiting outside the fraction in the original problem.

That gives us $x^3\cdot x^3=x^6$ in the numerator. In the denominator, we just extracted $x^6$.

The $x^6$ in the numerator and the $x^6$ in the denominator cancel out perfectly! The 'infinity' part of the problem has been neutralized.

We are left with a clean, finite expression where all the $\frac{1}{x}$ and $\frac{1}{x^2}$ terms simply vanish into zero as $x\to \infty$.

With the $x$ terms gone, the expression collapses into simple arithmetic. The numerator becomes $(\sqrt{3}+\sqrt{3}{)}^6$, which is $(2\sqrt{3}{)}^6$.

The denominator becomes $(1+1{)}^6$, which is $2^6$. Calculating this, we get:

The $2^6$ terms cancel out, leaving us with $(\sqrt{3}{)}^6$, which is $3^3$, or 27.

You see? By refusing to be intimidated by the complexity and instead focusing on the structural dominance of the terms, we turned a terrifying limit into a simple, beautiful number. Keep this perspective in your toolkit: in limits, always look for what truly matters as $x$ grows, and let the rest fade away.