Question

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

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

Answer 1

Answer: (A)

$\frac{3}{2}$

Answer 2

$\displaystyle\lim _{x \rightarrow 0} \, \frac{e^{x^{2}}-\cos x}{x^{2}}$

Using L'Hospital's rule,

$\lim_{ x\rightarrow 0} \frac{2xe^{x^2} + Sinx}{2x}$

$\lim_{a \rightarrow b} e^{x^2} + \lim_{a \rightarrow b} \frac{Sinx}{2x}$

$=1+\frac1{2} = \frac3{2}$

So, The correct option is A.