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Question: $\lim_{x\to\infty} 2^x(4\tan^{-1}(1+2^{-x})-\pi)$ is equal to...

lim⁡x→∞2x(4tan⁡−1(1+2−x)−π)\lim_{x\to\infty} 2^x(4\tan^{-1}(1+2^{-x})-\pi) is equal to

A

1

B

2

C

3

D

4

Answer

2

Explanation

Solution

Let y=2−xy = 2^{-x}. As x→∞x \to \infty, y→0y \to 0. The limit becomes lim⁡y→04tan⁡−1(1+y)−πy\lim_{y\to 0} \frac{4\tan^{-1}(1+y)-\pi}{y}. This is a 00\frac{0}{0} form. Using L'Hopital's rule, the limit is lim⁡y→041+(1+y)21=41+(1)2=2\lim_{y\to 0} \frac{\frac{4}{1+(1+y)^2}}{1} = \frac{4}{1+(1)^2} = 2.