+7 How would one find
\(\lim_{x\rightarrow 0} (\frac{(1+x)^{1/x}}{e})^{1/x}\)
using L'Hospital's rule?
+118658 Wolfram alpha gives a step by step solution here
tps://www.wolframalpha.com/input/?i=lim+as+x+tends+to+0+of+((1%2Bx)%5E(1%2Fx)%2Fe)%5E(1%2Fx)
+118658 The link didn't work, I will try again.
https://www.wolframalpha.com/input/?i=lim+as+x+tends+to+0+of+((1%2Bx)%5E(1%2Fx)%2Fe)%5E(1%2Fx)
+26388 How using L'Hospital's rule?
\(\displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} }\)
Formula: \(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{x\to 0} f(x)\right) &=& \displaystyle\lim \limits_{x\to 0} \ln f(x) \\ \hline \end{array} \)
1. We take the logarithm:
\(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} }\right) &=& \displaystyle\lim \limits_{x\to 0} \ln \left\{ \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} } \right\}\\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \ln \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)\\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \left[ \ln \left( (1+x)^{ \frac{1}{x} } \right) -\ln(e) \right] \quad &|\quad \ln(e) = 1\\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \left[ \ln \left( (1+x)^{ \frac{1}{x} } \right) -1 \right] \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \left[ \frac{1}{x}\cdot\ln(1+x) -1 \right] \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \left[ \frac{1}{x}\cdot\ln(1+x) -\frac{x}{x} \right] \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \left[ \frac{\ln(1+x)-x}{x} \right] \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{\ln(1+x)-x}{x^2} \quad &|\quad \text{L'Hospital's rule}\\ &=& \displaystyle\lim \limits_{x\to 0} \frac{(\ln(1+x)-x)'}{(x^2)'} \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{ \frac{1}{1+x}-1 }{2x} \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{ \frac{1-1-x}{1+x} }{2x} \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{ \frac{ -x}{1+x} }{2x} \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{ -x}{2x(1+x)} \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{ -1}{2(1+x)} \quad &|\quad x = 0\\ &=& \frac{ -1}{2(1+0)} \\ &=& - \frac{1}{2} \\ \hline \end{array} \)
2. We revert the logarithm:
\(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} }\right) &=& - \frac{1}{2} \\\\ e^{\ln \left(\displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} }\right) } &=& e^{- \frac{1}{2} } \\\\ \displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} } &=& e^{- \frac{1}{2} } \\ &=& \frac{1}{ e^{ \frac{1}{2} } } \\ \mathbf{ \displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} } } &\mathbf{=}& \mathbf{\frac{1}{ \sqrt{e} } } \\ \hline \end{array}\)
+7 This may be a slightly late response, but yes, that's correct. Thank you.
I did manage to determine the solution in the time that I was waiting for the answer, as the assignment was due the next day, but thank you nonetheless!
