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Question: Consider \(\frac{r^{3} - 8}{r^{3} + 8}\)...

Consider r3−8r3+8\frac{r^{3} - 8}{r^{3} + 8}

A

Limh→0Δp3\underset{h \rightarrow 0}{Lim}\frac{\Delta}{p^{3}} is discontinuous everywhere

B

lim⁡n→∞\lim_{n \rightarrow \infty} is continuous everywhere but not differentiable at

x = 0

C

∑r=1nr2n3\frac{\sum_{r = 1}^{n}r^{2}}{n^{3}}exists in (–1, 1)

D

lim⁡n→∞\lim_{n \rightarrow \infty}exists in (–2, 2)

Answer

lim⁡n→∞\lim_{n \rightarrow \infty} is continuous everywhere but not differentiable at

x = 0

Explanation

Solution

We have, f(x)={x2∣x∣,x≠00,x=0f ( x ) = \left\{ \begin{array} { l l } \frac { x ^ { 2 } } { | x | } , & x \neq 0 \\ 0 , & x = 0 \end{array} \right. =

⇒ lim⁡x→0−f(x)=lim⁡x→0−−x=0\lim _ { x \rightarrow 0 ^ { - } } f ( x ) = \lim _ { x \rightarrow 0 ^ { - } } - x = 0, lim⁡x→0+f(x)=lim⁡x→0+x=0\lim _ { x \rightarrow 0 ^ { + } } f ( x ) = \lim _ { x \rightarrow 0 ^ { + } } x = 0 and f(0)f ( 0 )

= 0.

So f(x)f ( x ) is continuous at x=0x = 0. Alsof(x)f ( x ) is continuous for all other values of x. Hence, f(x)f ( x ) is everywhere continuous.

Also, Rf′(0)=f′(0+0)R f ^ { \prime } ( 0 ) = f ^ { \prime } ( 0 + 0 )= lim⁡h→0f(h)−f(0)h−0\lim _ { h \rightarrow 0 } \frac { f ( h ) - f ( 0 ) } { h - 0 } =lim⁡h→0h−0h=1\lim _ { h \rightarrow 0 } \frac { h - 0 } { h } = 1

i.e. Rf′(0)=1R f ^ { \prime } ( 0 ) = 1 and Lf′(0)=f′(0−0)=lim⁡h→0f(−h)−f(0)−hL f ^ { \prime } ( 0 ) = f ^ { \prime } ( 0 - 0 ) = \lim _ { h \rightarrow 0 } \frac { f ( - h ) - f ( 0 ) } { - h }

=lim⁡h→0h−h=−1\lim _ { h \rightarrow 0 } \frac { h } { - h } = - 1

i.e. Lf′(0)=−1L f ^ { \prime } ( 0 ) = - 1 So, Lf′(0)≠Rf′(0)L f ^ { \prime } ( 0 ) \neq R f ^ { \prime } ( 0 )

i.e., f(x)f ( x ) is not differentiable at x=0x = 0.