Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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# If F ( X ) = X 2 + X 2 1 + X 2 + X 2 ( 1 + X 2 ) + . . . + X 2 ( 1 + X 2 ) + . . . . , Then at X = 0, F (X) (A) Has No Limit (B) is Discontinuous (C) is Continuous but Not Differentiable - Mathematics

#### Question

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

then at x = 0, f (x)

##### Options
• has no limit

• is discontinuous

• is continuous but not differentiable

• is differentiable

#### Solution

(b) is discontinuous

$\text{We have},$
$f\left( x \right) = x^2 + \frac{x^2}{1 + x^2} + \frac{x^2}{\left( 1 + x^2 \right)} + . . . + \frac{x^2}{\left( 1 + x^2 \right)} + . . . . ,$
$\text{When x = 0 then } x^2 = 0$
$\text { and } \frac{x^2}{1 + x^2} = 0$
$\therefore f\left( 0 \right) = 0 + 0 + 0 + 0 . . . . . . .$
$\Rightarrow f\left( 0 \right) = 0$
$\text { When, x } \neq 0$
$\text{Then,} x^2 > 0$
$\text { and }1 + x^2 > x^2$
$\Rightarrow 0 < \frac{x^2}{1 + x^2} < 1$
$\therefore \lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} \left( x^2 + \frac{x^2}{1 + x^2} + \frac{x^2}{\left( 1 + x^2 \right)} + . . . + \frac{x^2}{\left( 1 + x^2 \right)} + . . . . , \right)$
$= \lim_{x \to 0} \left[ x^2 \left( 1 + \frac{1}{1 + x^2} + \frac{1}{\left( 1 + x^2 \right)} + . . . + \frac{1}{\left( 1 + x^2 \right)} + . . . . , \right) \right]$
$= \lim_{x \to 0} \left[ x^2 \left( \frac{1}{1 - \frac{1}{1 + x^2}} \right) \right] \left[ \text{Sum of infinite series where}, r = \frac{1}{1 + x^2} \right]$
$= \lim_{x \to 0} \left[ x^2 \left( \frac{1 + x^2}{x^2} \right) \right]$
$= \lim_{x \to 0} \left( 1 + x^2 \right)$
$= 1$
$\therefore \lim_{x \to 0} f\left( x \right) \neq f\left( 0 \right)$
$\therefore f\left( x \right) \text { is discontinuous at } x = 0$

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