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

My dear student, welcome to the arena of limits. Today, we are going to dissect a problem that often intimidates students at first glance, but reveals itself to be a masterpiece of algebraic elegance once you understand its soul.

We are looking at the limit:

The Diagnosis

In the JEE Advanced examination, the first step is always diagnosis. Never rush into calculation.

Look at the base: $f(x)=\frac{3x^2+2}{7x^2+2}$. As $x$ approaches $0$, the numerator becomes $3(0{)}^2+2=2$, and the denominator becomes $7(0{)}^2+2=2$. Thus, the base approaches $\frac{2}{2}=1$.

Now, look at the exponent: $g(x)=\frac{1}{x^2}$. As $x$ approaches $0$, the denominator $x^2$ becomes an infinitesimally small positive number. Dividing $1$ by an infinitesimal number sends the value to infinity.

We have confirmed the indeterminate form $1^{\infty }$. This is not a number; it is a battle between the base approaching $1$ and the exponent approaching infinity. We need a strategy.

The Weapon of Choice

We do not use brute force here. We use the standard identity for $1^{\infty }$ limits.

Whenever you see ${\lim }_{x\to a}[f(x){]}^{g(x)}$ where $f(x)\to 1$ and $g(x)\to \infty$, you can immediately rewrite the limit as:

This formula is your best friend. It effectively 'linearizes' the exponent, turning a terrifying power-based problem into a simple multiplication problem. Let us apply this to our specific case.

The Algebraic Dance

Substituting our functions into the formula, we get:

Now, focus your attention solely on the bracketed term. We need to simplify $\frac{3x^2+2}{7x^2+2}-1$. To do this, we take the Least Common Multiple (LCM):

Be careful here! The negative sign must distribute across the entire numerator of the second term. This is where many students lose marks. When we distribute, we get:

The constant terms $2$ and $-2$ cancel out beautifully, leaving us with:

The Final Victory

Now, we reassemble our limit. We have:

Look at the beauty of this moment. We have an $x^2$ in the numerator (from our simplification) and an $x^2$ in the denominator (from our original exponent). Since $x\to 0$ but $x eq 0$, we can safely cancel these terms.

The indeterminate nature of the limit vanishes instantly:

Now, we simply substitute $x=0$. The denominator becomes $7(0{)}^2+2=2$. The exponent becomes $\frac{-4}{2}=-2$.

Our final answer is:

See how the complexity dissolved? We moved from a daunting exponential form to a simple algebraic fraction, and finally to a constant. This is the rhythm of calculus. Trust the process, respect the indeterminate forms, and the math will always reward you.