Question

lf $f\left(x\right)=\{\begin{array}{cc}\frac{1-\cos x}{x}& x\ne 0\\ 0& x=0\end{array}$, then $f^{{}^{\prime }}\left(0\right)=$

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{3}{4}$
• D. Does not exist

Answer 1

The correct option is A $\frac{1}{2}$

Using fundamental theorem,
$f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$

$=\underset{h\to 0}{lim}\frac{\frac{1-\cos h}{h}-0}{h}$

$=\underset{h\to 0}{lim}\frac{1-\cos h}{h^2}$

$=\underset{h\to 0}{lim}\frac{1}{2}.\frac{{\sin }^2\frac{h}{2}}{\frac{h^2}{4}}=\frac{1}{2}$