Question

$\underset{x\to 0}{lim}\frac{e^{x^2}-\cos x}{{\sin }^{2}x}$ is equal to :

• A. 2
• B. $\frac{3}{2}$
• C. 3
• D. $\frac{5}{4}$

Answer 1

Answer: (B)

$\frac{3}{2}$

$\underset{x\to 0}{lim}\frac{e^{x^2}-\cos x}{{\sin }^{2}x}$

$=\underset{x\to 0}{lim}\frac{\frac{e^{x^2}-\cos x}{x^2}}{{\left(\frac{\sin x}{x}\right)}^2}=\underset{x\to 0}{lim}\frac{\frac{1+x^2+\frac{x^4}{2!}-(1-\frac{x^2}{2!}+\frac{x^4}{4!}-..)}{x^2}}{{\left(\frac{\sin x}{x}\right)}^2}=1+\frac{1}{2!}=\frac{3}{2}$