Question

Calculate each of the limits in Exercises 21-48. Some of these limits are made easier by L'Hôpital's rule, and some are not.

$\underset{x\to 0^{+}}{\lim }\left(x^{-1}-\frac{1}{2^x-1}\right)$

Short answer

The value of limit is$\frac{\ln 2-1}{\ln 2}$

Step-by-step solution

Step 1. Given information

Given function$\underset{x\to 0^{+}}{\lim }\left(x^{-1}-\frac{1}{2^x-1}\right)$

Apply L-Hospital rule and solve

Calculating, we get

$$\begin{gathered} \underset{x\to 0^{+}}{\lim }\left(x^{-1}-\frac{1}{2^x-1}\right)=\underset{x\to 0^{+}}{\lim }\left(\frac{1}{x}-\frac{1}{2^x-1}\right) \\ \underset{x\to 0^{+}}{\lim }\left(x^{-1}-\frac{1}{2^x-1}\right)=\underset{x\to 0^{+}}{\lim }\left(\frac{2^x-1-x}{x\left(2^x-1\right)}\right) \\ \underset{x\to 0^{+}}{\lim }\left(x^{-1}-\frac{1}{2^x-1}\right)=\underset{x\to 0^{+}}{\lim }\left(\frac{2^x-1-x}{x·2^x-x}\right) \\ =\underset{x\to 0^{+}}{\lim }\left(\frac{2^x\ln 2-0-1}{x·2^x\ln 2+2^x·1-1}\right) \\ =\underset{x\to 0^{+}}{\lim }\left(\frac{2^x\ln 2-1}{x·2^x\ln 2+2^x-1}\right) \\ =\frac{2^0\ln 2-1}{0·2^0\ln 2+2^0-1} \\ =\frac{\ln 2-1}{\ln 2+1-1} \\ =\frac{\ln 2-1}{\ln 2} \end{gathered}$$