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प्रश्न
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)
पर्याय
has no limit
is discontinuous
is continuous but not differentiable
is differentiable
उत्तर
(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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