MCQ
If $\text{f(x)}=\text{x}^2+\frac{\text{x}^2}{1+\text{x}^2}+\frac{\text{x}^2}{(1+\text{x}^2)}+....+\frac{\text{x}^2}{(1+\text{x}^2)}+....,$ then at x = 0, f(x):
  • A
    Has not limit.
  • Is discontinuous.
  • C
    Is continuous but not differentiable.
  • D
    Is differentiable.

Answer

Correct option: B.
Is discontinuous.
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\Big(\text{x}^2+\frac{\text{x}^2}{1+\text{x}^2}+\frac{\text{x}^2}{(1+\text{x}^2)^2}+....+\frac{\text{x}^2}{(1+\text{x}^2)^\text{n}}+....,\Big)$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\text{x}^2\Big(1+\frac{\text{x}^2}{1+\text{x}^2}+\frac{\text{x}^2}{(1+\text{x}^2)^2}+....+\frac{\text{x}^2}{(1+\text{x}^2)^\text{n}}+....,\Big)$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\text{x}^2\bigg(\frac{1}{1-\frac{1}{1+\text{x}^2}}\bigg)$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}(1+\text{x}^2)$
$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}=1$
But, f(0)=0
$\text{f}(0)\neq\lim\limits_{\text{x}\rightarrow0}\text{f(x)}$
Function is discontinuous.

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