Solve this

Question:

If $f(x)=x^{2}+\frac{x^{2}}{1+x^{2}}+\frac{x^{2}}{\left(1+x^{2}\right)}+\ldots+\frac{x^{2}}{\left(1+x^{2}\right)}+\ldots$

then at $x=0, f(x)$

(a) has no limit

(b) is discontinuous

(c) is continuous but not differentiable

(d) is differentiable

Solution:

(b) is discontinuous

We have,

$f(x)=x^{2}+\frac{x^{2}}{1+x^{2}}+\frac{x^{2}}{\left(1+x^{2}\right)}+\ldots+\frac{x^{2}}{\left(1+x^{2}\right)}+\ldots$

When $x=0$ then $x^{2}=0$

and $\frac{x^{2}}{1+x^{2}}=0$

$\therefore f(0)=0+0+0+0 \ldots \ldots$

$\Rightarrow f(0)=0$

When, $x \neq 0$

Then, $x^{2}>0$

and $1+x^{2}>x^{2}$

$\Rightarrow 0<\frac{x^{2}}{1+x^{2}}<1$

$\therefore \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(x^{2}+\frac{x^{2}}{1+x^{2}}+\frac{x^{2}}{\left(1+x^{2}\right)}+\ldots+\frac{x^{2}}{\left(1+x^{2}\right)}+\ldots,\right$,

$=\lim _{x \rightarrow 0}\left[x^{2}\left(1+\frac{1}{1+x^{2}}+\frac{1}{\left(1+x^{2}\right)}+\ldots+\frac{1}{\left(1+x^{2}\right)}+\ldots,\right)\right$,

$=\lim _{x \rightarrow 0}\left[x^{2}\left(\frac{1}{1-\frac{1}{1+x^{2}}}\right)\right]$           $\left[\right.$ Sum of infinite series where, $\left.r=\frac{1}{1+x^{2}}\right]$

$=\lim _{x \rightarrow 0}\left[x^{2}\left(\frac{1+x^{2}}{x^{2}}\right)\right]$

$=\lim _{x \rightarrow 0}\left(1+x^{2}\right)$

$=1$

$\therefore \lim _{x \rightarrow 0} f(x) \neq f(0)$

$\therefore f(x)$ is discontinuous at $x=0$

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