$$\lim_{x \to 0} \left(\frac{3x^2+2}{7x^2+2}\right)^{\frac{1}{x^2}}$$ is equal to

We have to find the limit

$$\displaystyle \lim_{x \to 0}\left(\frac{3x^{2}+2}{7x^{2}+2}\right)^{\tfrac{1}{x^{2}}}.$$

Whenever a limit is of the indeterminate form $$1^{\infty},$$ a standard method is to write the inner expression in the form $$1+u(x)$$ where $$u(x)\to 0$$ as $$x\to 0,$$ and then recall the well-known formula

$$\lim_{u\to 0}\left(1+u\right)^{\tfrac{1}{u}} = e.$$

For our problem, let us define

$$A(x)=\frac{3x^{2}+2}{7x^{2}+2}.$$

First we check the value of $$A(x)$$ at $$x=0$$:

$$A(0)=\frac{3\cdot 0^{2}+2}{7\cdot 0^{2}+2}=\frac{2}{2}=1.$$

So indeed $$A(x)\to 1$$ as $$x\to 0,$$ and the whole expression is of the desired $$1^{\infty}$$ type. Now we write

$$A(x)=1+u(x),\quad\text{where}\quad u(x)=\frac{3x^{2}+2}{7x^{2}+2}-1.$$

We simplify $$u(x)$$ step by step:

$$u(x)=\frac{3x^{2}+2-(7x^{2}+2)}{7x^{2}+2}.$$

Combining like terms in the numerator, we obtain

$$u(x)=\frac{3x^{2}+2-7x^{2}-2}{7x^{2}+2} =\frac{-4x^{2}}{7x^{2}+2}.$$

Clearly, $$u(x)\to 0$$ as $$x\to 0,$$ so the expression

$$\left(1+u(x)\right)^{\tfrac{1}{x^{2}}}$$

is suitable for the exponential trick. The limit we seek can be rewritten as

$$\lim_{x\to 0}\left(1+u(x)\right)^{\tfrac{1}{x^{2}}} =\exp\!\Bigl(\,\lim_{x\to 0}\tfrac{u(x)}{x^{2}}\Bigr).$$

Thus we must compute

$$\lim_{x\to 0}\frac{u(x)}{x^{2}} =\lim_{x\to 0}\frac{-4x^{2}}{x^{2}\bigl(7x^{2}+2\bigr)}.$$

The factor $$x^{2}$$ cancels:

$$\frac{u(x)}{x^{2}} =\frac{-4}{7x^{2}+2}.$$

Now let $$x\to 0$$ in the remaining expression:

$$\lim_{x\to 0}\frac{-4}{7x^{2}+2} =\frac{-4}{2} =-2.$$

Substituting this result back into the exponential, we get

$$\exp\!\left(-2\right)=e^{-2}=\frac{1}{e^{2}}.$$

Hence, the correct answer is Option B.