Question

If $f\left(x\right)=\frac{x-e^x+\cos 2x}{x^2}$,$x\ne 0$

• A. $f\left(0\right)=\frac{5}{2}$
• B. $f\left(0\right)=-2$
• C. $f\left(0\right)=-0.5$
• D. $f\left(0\right)=-2.5$

Answer 1

Answer: (D)

$f\left(0\right)=-2.5$

$\underset{x\to 0}{lim}\frac{x-e^x+\cos 2x}{x^2}$
Applying L'Hospital's rule
$=\underset{x\to 0}{lim}-\frac{-1+e^x+2\sin 2x}{2x}$
Again applying L'Hospital's rule
$=\underset{x\to 0}{lim}\frac{-e^x-4\cos 2x}{2}=-\frac{5}{2}$