AP EAMCET · Maths · Continuity and Differentiability
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 \(\mathrm{x}=\)
- A 0
- B \(\frac{1}{3}\)
- C 1
- D -1
Answer & Solution
Correct Answer
(B) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{S}_{\mathrm{n}}=1+3 \mathrm{x}+9 \mathrm{x}^2+27 \mathrm{x}^3+\ldots \mathrm{n}\) terms \[ \begin{aligned} & \lim _{n \rightarrow \infty} S_n=1+3 x+9 x^2+27 x^3+\ldots \\ & \Rightarrow \lim _{n \rightarrow \infty} S_n=(1-3 x)^{-1}=\frac{1}{1-3 x} \end{aligned} \] So,…
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