Let $S_n=1+3 x+9 x^2+27 x^3+\ldots n$ terms and $-\frac{1}{3}

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Let $S_n=1+3 x+9 x^2+27 x^3+\ldots n$ terms and $-\frac{1}{3} < x < \frac{1}{3}$ If $\lim _{\mathrm{n} \rightarrow \infty} \mathrm{S}_{\mathrm{n}}=\mathrm{f}(\mathrm{x})$, then $\mathrm{f}(\mathrm{x})$ is discontinuous at the point $\mathrm{x}=$

Aicharya · Accepted Answer

$\mathrm{S}_{\mathrm{n}}=1+3 \mathrm{x}+9 \mathrm{x}^2+27 \mathrm{x}^3+\ldots \mathrm{n}$ terms $ \begin{aligned} & \lim _{n ightarrow \infty} S_n=1+3 x+9 x^2+27 x^3+\ldots \ & \Rightarrow \lim _{n ightarrow \infty} S_n=(1-3 x)^{-1}=\frac{1}{1-3 x} \end{aligned} $ So, $f(x)=\frac{1}{1-3 x}$ $\mathrm{f}(\mathrm{x})$ is discontinuous on $\mathrm{x}=\frac{1}{3}$

Let $S_n=1+3 x+9 x^2+27 x^3+\ldots n$ terms and $-\frac{1}{3} < x < \frac{1}{3}$ If $\lim _{\mathrm{n}…

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