$$\lim _{t\to 0}\left(\int _t^{2t}\:\left(\frac{e^{2x}-1}{x^2}\right)dx\right)$$
So i got this question on my math exam today and I had no idea how to solve it. I looked at all my notes and tried looking for a solution online but couldn't find anything. Any help would be appreciated.
Expand $e^{2x}$ as:
$$e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + ... $$
$$\frac{e^{2x}-1}{x^2} = \frac{2}{x} + \frac{2^2}{2!} + \frac{2^3 x}{3!} + ... $$
Integrating the above expression and applying the limits gives
$$2 \ln(2)$$
as the answer.
Have left out the details for you to work them out.
Cheers!
$$e^{2x}-1=\sum_{k=0}^\infty\frac{(2x)^k}{k!}-1=\sum_{k=1}^\infty\frac{(2x)^k}{k!}$$ $$\begin{align} \int _t^{2t}\left(\frac{e^{2x}-1}{x^2}\right)dx &=\int _t^{2t}\left(\frac1{x^2}\sum_{k=1}^\infty\frac{(2x)^k}{k!}\right)dx\\ &=\sum_{k=1}^\infty\left(\frac{2^k}{k!}\int _t^{2t}x^{k-2}dx\right)\\ &=2\int _t^{2t}\frac1xdx+\sum_{k=2}^\infty\left(\frac{2^k}{k!}\int _t^{2t}x^{k-2}dx\right)\\ &=2[\ln{|x|}]_t^{2t}+\sum_{k=2}^\infty\left(\frac{2^k}{k!}\left[\frac1{k-1}x^{k-1}\right]_t^{2t}\right)\\ &=2\ln{|2t|}-2\ln{|t|}+\sum_{k=2}^\infty\left(\frac{2^k}{k!}\cdot\frac{(2t)^{k-1}-t^{k-1}}{k-1}\right)\\ &=2\ln{(2)}+\sum_{k=2}^\infty\left(\frac{(2t)^{k-1}}{k!}\cdot\frac{2^k-2}{k-1}\right)\\ &=2\ln{(2)}+\sum_{k=1}^\infty\left(\frac{(2t)^k}{(k+1)!}\cdot\frac{2^{k+1}-2}{k}\right)\\ &=2\ln{(2)}+2\sum_{k=1}^\infty\left(\frac{(4t)^k}{k\cdot(k+1)!}\right)-2\sum_{k=1}^\infty\left(\frac{(2t)^k}{k\cdot(k+1)!}\right)\\ \end{align}$$ Now taking the limit as $t\to0$ $$\begin{align} \lim_{t\to0}\left(\int _t^{2t}\left(\frac{e^{2x}-1}{x^2}\right)dx\right) &=\lim_{t\to0}\left(2\ln{(2)}+2\sum_{k=1}^\infty\left(\frac{(4t)^k}{k\cdot(k+1)!}\right)-2\sum_{k=1}^\infty\left(\frac{(2t)^k}{k\cdot(k+1)!}\right)\right)\\ &=\boxed{2\ln{(2)}}\\ \end{align}$$
