Question

The value of $\underset{x\to 0}{lim}\frac{e^{2x^2}-1}{x^2}$ using Maclaurin series is?

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

${lim}_{x\to 0}\frac{e^{2x^2}-1}{x^2}=?$
We have to find the limit value using Maclaurin series.
The Maclaurin series of $e^{2x^2}$ is $1+2x^2+2x^4$.
The Maclaurin series of $x^2$ is $x^2$.
${lim}_{x\to 0}\frac{e^{2x^2}-1}{x^2}={lim}_{x\to 0}\frac{1+2x^2+2x^4-1}{x^2}$
$={lim}_{x\to 0}\frac{2x^2+2x^4}{x^2}$
$={lim}_{x\to 0}\frac{2x^2(1+x^2)}{x^2}$
$={lim}_{x\to 0}2(1+x^2)$
${lim}_{x\to 0}\frac{e^{2x^2}-1}{x^2}=2$