Question

Applying Lagrange's mean value theorem for a suitable function $f\left(x\right)$ in $\left[0,h\right]$, we have $f\left(h\right)=f\left(0\right)+hf\text{'}\left(\theta h\right),0<\theta <1$. Then, for $f\left(x\right)=\cos x$, the value of $li{m}_{h\to 0+}\theta$ is

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

Answer 1

The correct option is C

$\frac{1}{2}$

Explanation for the correct option:

Step 1: Use Lagrange's mean value theorem to write the value of $\cos h$:

Given: $f\left(h\right)=f\left(0\right)+hf\text{'}\left(\theta h\right),0<\theta <1$

Put $f\left(h\right)=\cos h$ in the equation.

$$\begin{gathered} \cos h=1+h(-\sin (\theta h\left)\right) \\ \Rightarrow \sin \left(\theta h\right)=\frac{\left[1-\cos h\right]}{h} \\ \Rightarrow \left(\theta h\right)={\sin }^{-1}\left(\frac{\left[1-\cos h\right]}{h}\right) \\ \Rightarrow \theta =\frac{{\sin }^{-1}\left(\frac{\left[1-\cos h\right]}{h}\right)}{h} \end{gathered}$$

Step 2: Apply limit on both sides:

Put $\underset{h\to 0^{+}}{\lim }$ on both sides of the equation.

$\begin{array}{rcl}\underset{h\to 0^{+}}{\lim }\theta & =& \underset{h\to 0^{+}}{\lim }\frac{{\sin }^{-1}\left(\frac{\left[1-\cos h\right]}{h}\right)}{h}\\ & =& \underset{h\to 0^{+}}{\lim }\frac{{\sin }^{-1}\left({\displaystyle \frac{h}{2}}\right)}{{\displaystyle \frac{h}{2}}}\times \frac{1}{2}\\ & =& \frac{1}{2}\end{array}$

Hence, option (C) is the correct answer.