Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to 0}{\left(\frac{1+5x^2}{1+3x^2}\right)}^{1/x^2}=...$

Conquering the $1^{\infty }$ Indeterminate Form

Welcome, future engineer. Today, we are going to peel back the layers of a problem that often intimidates students at first glance. We are looking at the limit:

At first, it looks like a mess of variables and fractions. But in the world of JEE Advanced, we don't fear complexity; we dismantle it. Let's embark on this journey together.

The Diagnosis

In calculus, the first rule of engagement is always direct substitution. We need to know what we are dealing with. Let's look at the base:

As $x$ approaches zero, the numerator becomes $1$ and the denominator becomes $1$. So, the base is approaching $1$.

Now, look at the exponent: $\frac{1}{x^2}$. As $x$ gets infinitesimally small, $x^2$ becomes a tiny positive number, and $1$ divided by a tiny number shoots off to infinity. We have arrived at the classic $1^{\infty }$ indeterminate form. This is not a number; it is a signal that we need a specific tool to resolve the tension between the base and the exponent.

The Tool of Choice

We cannot simply plug in zero here. We need the $e^L$ property. This is a beautiful piece of mathematical machinery.

It tells us that if we have a limit of the form ${\lim }_{x\to a}[f(x){]}^{g(x)}$ where $f(x)\to 1$ and $g(x)\to \infty$, we can rewrite the entire expression as:

Think of this as a transformation that brings the exponent down to earth, turning a power problem into a multiplication problem. It is elegant, efficient, and exactly what we need.

The Algebraic Dance

Now, let's map our problem to this formula. Our base function is $f(x)=\frac{1+5x^2}{1+3x^2}$, and our exponent function is $g(x)=\frac{1}{x^2}$. Substituting these into our formula gives us:

I know this looks intimidating, but let's take a breath and focus only on the bracketed term: $\frac{1+5x^2}{1+3x^2}-1$. To combine these, we need a common denominator, which is $1+3x^2$. This gives us:

Watch closely as we expand the numerator: $1+5x^2-1-3x^2$. The constants $1$ and $-1$ vanish! This is the moment of clarity we were waiting for. We are left with $5x^2-3x^2$, which simplifies to $2x^2$. Our bracket is now simply $\frac{2x^2}{1+3x^2}$.

The Victory

We are almost there. Let's bring back the exponent function $g(x)=\frac{1}{x^2}$ and multiply it by our simplified bracket:

Look at that! The $x^2$ in the numerator and the $x^2$ in the denominator cancel out perfectly. We are left with $\frac{2}{1+3x^2}$.

With the indeterminacy removed, we can finally perform direct substitution. As $x\to 0$, the term $3x^2$ becomes $0$. We are left with $\frac{2}{1+0}=2$.

Remember, this is the value of the exponent of $e$. Therefore, our final answer is $e^2$. See how the complexity dissolved? We didn't need brute force; we needed the right perspective. Keep this logic in your toolkit, and no limit will ever stand in your way.