Question

Let $f(x)=x^3.$ Use mean value theorem to write $\frac{f(x+h)-f(x)}{h}=f^{\prime }(x+\theta h)$ , with $0<\theta <1$ . If $x\ne 0,$ then $\underset{h\to 0}{\lim }\theta$ is equal to

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

Answer 1

Answer: (C)

Given, $f(x)=x^3$
$\therefore$ $f(x+h)={(x+h)}^3$
Now, $f^{\prime }(x)=3x^2$
$\therefore$ $f^{\prime }(x+\theta h)=3{(x+\theta h)}^2$
Given, $\frac{f(x+h)-f(x)}{h}=f^{\prime }(x+\theta h)$
$\Rightarrow$ $\frac{{(x+h)}^3-x^3}{h}=3{(x+\theta h)}^2$
$\Rightarrow$ $\frac{x^3+h^3+3xh(x+h)-x^3}{h}$
$=3(x^2+{\theta }^2{h}^2+2x\theta h)$
$\Rightarrow$ $h^2+3x^2+3xh=3x^2+3{\theta }^2{h}^2+6x\theta h$
$\Rightarrow$ $3x=3{\theta }^{2}h+6x\theta$
Taking limit on both sides, we get
$\underset{h\to 0}{\lim }(h+3x)=\underset{h\to 0}{\lim }(3{\theta }^{2}h+6x\theta )$
$\Rightarrow$ $3x=6\underset{h\to 0}{\lim }\theta$
$\Rightarrow$ $\underset{h\to 0}{\lim }\theta =\frac{1}{2}=0.5$