1. We need to evaluate

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

2. This is of the standard form $1^{\infty}$, because as $x\to 0$,

$$\frac{3x^2+2}{7x^2+2}\to \frac{2}{2}=1,$$

while

$$\frac{1}{x^2}\to \infty.$$

So we take logarithm.

Let

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

Then

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

3. Simplify the fraction inside the logarithm:

$$\frac{3x^2+2}{7x^2+2}=\frac{2\left(1+\frac{3}{2}x^2\right)}{2\left(1+\frac{7}{2}x^2\right)} =\frac{1+\frac{3}{2}x^2}{1+\frac{7}{2}x^2}.$$

Hence

$$\ln\left(\frac{3x^2+2}{7x^2+2}\right) =\ln\left(1+\frac{3}{2}x^2\right)-\ln\left(1+\frac{7}{2}x^2\right).$$

4. Use the expansion

$$\ln(1+t)=t+o(t) \quad (t\to 0).$$

Therefore,

$$\ln\left(1+\frac{3}{2}x^2\right)=\frac{3}{2}x^2+o(x^2),$$

and

$$\ln\left(1+\frac{7}{2}x^2\right)=\frac{7}{2}x^2+o(x^2).$$

So,

$$\ln\left(\frac{3x^2+2}{7x^2+2}\right) =\left(\frac{3}{2}-\frac{7}{2}\right)x^2+o(x^2) =-2x^2+o(x^2).$$

5. Divide by $x^2$:

$$\ln L=\lim_{x\to 0}\frac{-2x^2+o(x^2)}{x^2}=-2.$$

Thus,

$$L=e^{-2}=\frac{1}{e^2}.$$

6. Option check:

• A: $e$ ❌
• B: $e^2$ ❌
• C: $\dfrac{1}{e^2}$ ✅
• D: $\dfrac{1}{e}$ ❌

Therefore, the correct answer is

$$\boxed{\frac{1}{e^2}}.$$