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AP EAMCET · Maths · Limits

If \(\lim _{n \rightarrow \infty} \frac{1-(10)^n}{1+(10)^{n+1}}=\frac{-\alpha}{10}\), then \(\alpha\) is equal to

  1. A 0
  2. B -1
  3. C 1
  4. D 2
Verified Solution

Answer & Solution

Correct Answer

(C) 1

Step-by-step Solution

Detailed explanation

\begin{aligned} & \lim _{n \rightarrow \infty} \frac{1-10^n}{1+10^{n+1}}=-\frac{\alpha}{10} \\ \Rightarrow & \lim _{n \rightarrow \infty} \frac{1-10^n}{1+10 \cdot 10^n}=-\frac{\alpha}{10} \\ & \lim _{n \rightarrow \infty}…