Question

Applying Lagranges's MeanValue Theorem for a suitable function $f\left(x\right)$ in $[0,h]$, we have $f\left(h\right)+h{f}^{\prime }\left(\theta h\right)$, $0<\theta <1$. Then for $f\left(x\right)=\cos x$, the value of $\underset{h\to 0^{+}}{lim}\theta$ is

• A. $1$
• B. $0$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

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

Note that

$\frac{f(a+h)-f\left(a\right)-h{f}^{\prime }\left(a\right)}{h^2}=\frac{f^{\prime }(a+\Theta h)-f^{\prime }\left(a\right)h}{h^2}$

$=\frac{f^{\prime }(a+\Theta h)-f^{\prime }\left(a\right)}{\Theta h}\Theta$

${lim}_{h\to 0}\frac{f^{\prime }(a+h)-f^{\prime }\left(a\right)}{2h}=\frac{1}{2}{f}^{\prime \prime }\left(a\right)=\frac{1}{2}{f}^{\prime \prime }\left(a\right){lim}_{h\to \Theta }\Theta$

Thus,

${lim}_{h\to o}\Theta =\frac{1}{2}$ is the correct answer.