Question

Let $f\left(x\right)=x^3$. Use the mean value theorem to write $\frac{\left[f(x+h)-f\left(x\right)\right]}{h}=f\text{'}(x+\theta h)$ with $0<\theta <1$. If $x\ne 0,$then lim$h_{h\to 0}\theta =$ ?

• A. $-1$
• B. $-0.5$
• C. $0.5$
• D. $1$

Answer 1

The correct option is C

$0.5$

Explanation for the correct option:

Given that: $f\left(x\right)=x^3$

Therefore, $f(x+h)=f\left(x\right)+hf\text{'}\left(x\right)+\left(\frac{h^2}{2}\right)f\text{'}\text{'}(x+{\theta }_{1}h)$, where $0<\theta <1$

Applying Lagrange's theorem,

$$\begin{gathered} \left[f(x+h)-f\left(x\right)\right]=hf\text{'}(x+\theta h),0<\theta <1 \\ f\text{'}(x+\theta h)-f\text{'}\left(x\right)={\theta }_{1}hf\text{'}\text{'}(x+{\theta }_{2}h) \\ \theta hf\text{'}\text{'}(x+{\theta }_2\theta h)=\frac{h}{2}f\text{'}(x+{\theta }_{1}h) \\ \theta =\frac{{\displaystyle \frac{1}{2}}\left[f\text{'}\text{'}(x+{\theta }_{1}h)\right]}{f\text{'}(x+{\theta }_{2}h)} \end{gathered}$$

now,

$$\begin{gathered} =\underset{h\to 0}{\lim }\theta \\ =\frac{1}{2}\left[\frac{f\text{'}\text{'}\left(x\right)}{f\text{'}\text{'}\left(x\right)}\right] \\ =\frac{1}{2} \end{gathered}$$

$f\text{'}\text{'}$ is continuous in $\left[x,x+h\right]$

Hence, the correct option is (C).