Question

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

• A. $e$
• B. $\frac{1}{e^{2}}$
• C. $\frac{1}{e}$
• D. $e^{2}$

Answer 1

The Correct Option is B

To find the limit of the given expression as \(x \to 0\), we need to evaluate:

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

This is an indeterminate form, which suggests using logarithms to simplify the expression. Let's redefine the expression as \(y\):

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

Taking the natural logarithm on both sides, we have:

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

We aim to find:

\lim_{x \to 0} \ln y.

Since, by L'Hopital's Rule, the limit of a logarithmic function can simplify the expression, compute:

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

Using Taylor expansion around \(x = 0\), we have the following expansions:

\ln(3x^2 + 2) \approx \ln 2 + \frac{3x^2}{2},

\ln(7x^2 + 2) \approx \ln 2 + \frac{7x^2}{2}.

Thus,

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

Substitute back into:

\ln y = \frac{1}{x^2} (-2x^2) = -2.

Therefore,

\lim_{x \to 0} \ln y = -2.

Thus, \(y\) can be expressed as:

y = e^{-2}.

This implies the final result of the original expression limit is:

\lim_{x \to 0} y = \frac{1}{e^2}.

Hence, the correct answer is \frac{1}{e^{2}} .