Question

The function $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}1-2x+3x^2-4x^3+....\infty ,& x\ne -1\end{array}\\ \begin{array}{cc}1,& x=-1\end{array}\end{array}$ is

• A. Continuous at $x=-1$
• B. Neither continuous nor derivable at $x=-1$
• C. Derivable at $x=-1$
• D. Not derivable at $x=-1$

Answer 1

Answer: (B), (D)

The correct options are
B Neither continuous nor derivable at $x=-1$
D Not derivable at $x=-1$

For $x\ne -1,$ we have

$f\left(x\right)=1-2x+3x^2-4x^3+...\infty ={(1+x)}^{-1}=\frac{1}{1+x}$

$\underset{h\to 0}{lim}f(-1-h)=\underset{h\to 0}{lim}\frac{1}{1-1-h}\to -\infty$

So, $f\left(x\right)$ is not continuous at $x=-1$

Also, $\underset{h\to 0}{lim}\frac{f(-1-h)-f(-1)}{(-1-h)-(-1)}=\underset{h\to 0}{lim}\frac{\frac{-1}{h}-1}{-h}=\underset{h\to 0}{lim}\frac{1+h}{h^2}\to \infty$

So, $f\left(x\right)$ is not derivable at $x=-1.$

Hence, $f\left(x\right)$ is neither continuous nor derivable at $x=-1$