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.

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

Short answer

The value of limit is$0$

Step-by-step solution

Step 1. Given information

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

Apply L-Hospital rule and solve

Calculating, we get

$$\begin{gathered} \underset{x\to 0^{+}}{\lim }\left(\frac{2^x-1}{x^2}-\frac{1}{1-e^x}\right)=\underset{x\to 0^{+}}{\lim }\left(\frac{\left(2^x-1\right)\left(1-e^x\right)-x^2}{x^2\left(1-e^x\right)}\right) \\ \underset{x\to 0^{+}}{\lim }\left(\frac{2^x-1}{x^2}-\frac{1}{1-e^x}\right)=\underset{x\to 0^{+}}{\lim }\left(\frac{2^x-2^x{e}^x-1+e^x-x^2}{x^2-x^2{e}^x}\right) \\ =\underset{x\to 0^{+}}{\lim }\left(\frac{2^x\ln 2-2^x{e}^x-e^x·2^x\ln 2+e^x-2x}{2x-x^2{e}^x-2x{e}^x}\right) \\ =\underset{x\to 0^{+}}{\lim }\left(\frac{2^x\ln 2-2^x{e}^x(1+\ln 2)+e^x-2x}{2x-x^2{e}^x-2x{e}^x}\right) \\ =\underset{x\to 0^{+}}{\lim }\left(\frac{2^x(\ln 2{)}^2-(1+\ln 2)\left(2^x{e}^x+e^x·2^x\ln 2\right)+e^x-2}{2-\left(x^2{e}^x+2x{e}^x\right)-2\left(x{e}^x+e^x·1\right)}\right) \\ =\underset{x\to 0^{+}}{\lim }\left(\frac{2^x(\ln 2{)}^2-(1+\ln 2)\left(2^x{e}^x+e^x·2^x\ln 2\right)+e^x-2}{2-x^2{e}^x-2x{e}^x-2x{e}^x-2e^x}\right) \\ =\underset{x\to 0^{+}}{\lim }\left(\frac{2^x(\ln 2{)}^2-(1+\ln 2)\left(2^x{e}^x+e^x·2^x\ln 2\right)+e^x-2}{2-x^2{e}^x-4x{e}^x-2e^x}\right) \\ =\underset{x\to 0^{+}}{\lim }\left(\frac{2^x(\ln 2{)}^2-(1+\ln 2{)}^2{2}^x{e}^x+e^x-2}{2-x^2{e}^x-4x{e}^x-2e^x}\right) \\ =\underset{x\to 0^{+}}{\lim }\left(\frac{2^x(\ln 2{)}^3-(1+\ln 2{)}^2\left(2^x{e}^x+e^x·2^x\ln 2\right)+e^x-0}{0-\left(2x{e}^x+x^2{e}^x\right)-4\left(x{e}^x+e^x·1\right)-2e^x}\right) \\ =\underset{x\to 0^{+}}{\lim }\left(\frac{2^x(\ln 2{)}^3-(1+\ln 2{)}^2\left(2^x{e}^x+e^x·2^x\ln 2\right)+e^x}{-\left(2x{e}^x+x^2{e}^x\right)-4\left(x{e}^x+e^x\right)-2e^x}\right) \\ =\frac{2^0(\ln 2{)}^3-(1+\ln 2{)}^2\left(2^0{e}^0+e^0·2^0\ln 2\right)+e^0}{-\left(2·0·e^0+0^2·e^0\right)-4\left(0·e^0+e^0\right)-2e^0} \\ =\frac{(\ln 2{)}^3-(1+\ln 2{)}^2(1+\ln 2)+1}{-4-2} \\ =\frac{(1+\ln 2{)}^3-(\ln 2{)}^3-1}{6} \\ =\frac{1^3+3·1^2·\ln 2+3(\ln 2{)}^2·1+(\ln 2{)}^3-(\ln 2{)}^3-1}{6} \\ =\frac{1+3\ln 2+3(\ln 2{)}^2+(\ln 2{)}^3-(\ln 2{)}^3-1}{6} \\ =\frac{3\ln 2+3(\ln 2{)}^2}{6} \\ =\frac{3\ln 2+3(\ln 2{)}^2}{6} \\ =\frac{(1+\ln 2)\ln 2}{2} \end{gathered}$$