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If $f(x)=x+\frac{x}{1+x}+\frac{x}{(1+x)^2}+\ldots$ to $\infty$, then at $\mathrm{x}=0, \mathrm{f}(\mathrm{x})$
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The correct answer is:
is discontinuous
For $\mathrm{x} \neq 0$, we have,
$$
f(x)=x+\frac{x / 1+x}{1-\frac{1}{1+x}}=x+\frac{x / 1+x}{x / 1+x}=x+1
$$
For $x=0, f(x)=0$. Thus, $f(x)=\left\{\begin{array}{cc}x+1, & x \neq 0 \\ 0, & x=0\end{array}\right.$ Clearly, $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=1 \neq f(0)$.
So, $\mathrm{f}(\mathrm{x})$ is discontinuous and hence not differentiable at $\mathrm{x}=0$.
$$
f(x)=x+\frac{x / 1+x}{1-\frac{1}{1+x}}=x+\frac{x / 1+x}{x / 1+x}=x+1
$$
For $x=0, f(x)=0$. Thus, $f(x)=\left\{\begin{array}{cc}x+1, & x \neq 0 \\ 0, & x=0\end{array}\right.$ Clearly, $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=1 \neq f(0)$.
So, $\mathrm{f}(\mathrm{x})$ is discontinuous and hence not differentiable at $\mathrm{x}=0$.
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