Question

$\displaystyle\lim_{x\to0}\frac{e^{x^2} -cos x}{x^{2}}=$

• A. $\frac{3}{2}$
• B. $\frac{1}{2}$
• C. 1
• D. $-\frac{3}{2}$
• E. $-\frac{1}{2}$

Answer 1

Answer: (C)

$ \displaystyle\lim _{x \rightarrow 0} \, \frac{e^{x^{2}}-\cos x}{x^{2}} $
$\left(1+\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}+\ldots \infty\right) $
$=\displaystyle\lim _{x \rightarrow 0} \frac{-\left(1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\ldots \infty\right)}{x^{2}} $
$=\displaystyle\lim _{x \rightarrow 0} \frac{x^{2}+\left(\frac{1}{2 !}-\frac{1}{4 !}\right) x^{4}+\ldots}{x^{2}} $
$= \displaystyle\lim _{x \rightarrow 0} 1+\left(\frac{1}{2}-\frac{1}{24}\right) x^{2}+\ldots=1+0 \ldots=1 $