Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to 0}(\frac{3x^2+2}{7x^2+2}{)}^{\frac{1}{x^2}}$ is equal to :

Imagine you are standing on the edge of a mathematical cliff, staring down at a function that seems to defy simple arithmetic. We are looking at the limit:

As $x$ dances closer to zero, the base $\frac{3x^2+2}{7x^2+2}$ settles down to $\frac{2}{2}$, which is $1$. Meanwhile, the exponent $\frac{1}{x^2}$ is racing toward infinity.

This is the classic $1^{\infty }$ indeterminate form—a tug-of-war where the base approaches $1$ while the exponent pulls the value toward infinity.

In the world of JEE Advanced, we use the exponential transformation rule to resolve this form:

Here, we identify our base as $f(x)=\frac{3x^2+2}{7x^2+2}$ and our exponent as $g(x)=\frac{1}{x^2}$. Plugging these into our formula, we get:

Now, we simplify the expression inside the bracket by finding a common denominator:

When we expand the numerator, the constants $2$ and $-2$ vanish, leaving us with $3x^2-7x^2=-4x^2$. Our expression now simplifies to:

We then reintroduce the exponent $\frac{1}{x^2}$ that was waiting outside the limit:

The $x^2$ terms in the numerator and the denominator cancel out perfectly. Since $x\to 0$ but $x eq 0$, this operation is valid:

Now, direct substitution is a breeze. As $x\to 0$, the term $7x^2$ becomes $0$, leaving us with:

You have successfully tamed the $1^{\infty }$ beast. The final answer is $e^{-2}$.