Question
If $$f\left( x \right) = x + \frac{x}{{1 + x}} + \frac{x}{{{{\left( {1 + x} \right)}^2}}} + .....{\text{to }}\infty ,$$ then at $$x = 0,\,f\left( x \right)$$
A.
has no limit
B.
is discontinuous
C.
is continuous but not differentiable
D.
is differentiable
Answer :
is discontinuous
Solution :
$$\eqalign{
& {\text{For }}x \ne 0,{\text{ we have}} \cr
& f\left( x \right) = x + \frac{{\frac{x}{{1 + x}} + x}}{{1 - \frac{1}{{1 + x}}}} = x + \frac{{\frac{x}{{1 + x}}}}{{\frac{x}{{1 + x}}}} = x + 1 \cr
& {\text{For }}x = 0,\,f\left( x \right) = 0 \cr} $$
Thus, \[f\left( x \right) = \left\{ \begin{array}{l}
x + 1,\,\,\,\,\,x \ne 0\\
\,\,\,\,\,0,\,\,\,\,\,\,\,\,\,\,\,\,x = 0
\end{array} \right.\]
$${\text{Clearly, }}\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 1 \ne f\left( 0 \right)$$
So, $$f\left( x \right)$$ is discontinuous and hence not differentiable at $$x = 0.$$