Is there any error in my proof of '$\lim_{x\to\infty} ((\frac{x}{x+k})^x)=\frac{1}{e^k}$'

Question

$\lim_{x\to\infty} ((\frac{x}{x+k})^x)$ $= \lim_{x\to\infty} (1-\frac{k}{x+k})^{x+x^k}\cdot(1-\frac{k}{x+k})^{-(x^k)} \\= \frac{\lim_{x\to\infty}(1-\frac{k}{x+k})^{x+x^k}}{\lim_{x\to\infty}(1-\frac{k}{x+k})^{x^k}} \\ = \frac{\lim_{x\to\infty}(1)^{x+x^k}}{\lim_{x\to\infty}(1-\frac{k}{x+k})^{x^k}} = \frac{1}{\lim_{x\to\infty}(1-\frac{k}{x+k})^{x^k}}\\= \frac{1}{\lim_{x\to\infty}(1+\frac{1}{x}-\frac{1}{x}-\frac{k}{x+k})^{x^k}} = \frac{1}{\lim_{x\to\infty}(1+\frac{1}{x})^{x^k}} = \frac{1}{\lim_{x\to\infty}((1+\frac{1}{x})^x)^k} = \frac{1}{e^k}$

Answer 1

Here is a valid solution: $x \ln (\frac x {x+k})=x\ln (1-\frac k {x+k}) \sim -x(\frac k {x+k}) \to -k$. Hence the limit is $e^{- k}$.

I have used the fact that $\frac {\ln {(1-x)}} x \to -1$ as $x \to 0$.

Answer 2

I do not believe this works. I would try rewriting it as follows though: $$\lim_{x\rightarrow\infty}(\frac{x}{x+k})^x=\lim_{x\rightarrow\infty}e^{\ln((\frac{x}{x+k})^x)}$$ $$=\lim_{x\rightarrow\infty}e^{x\ln(\frac{x}{x+k})}$$ $$=e^{\lim_{x\rightarrow\infty}x\ln(\frac{x}{x+k})}$$ $$=e^{\lim_{x\rightarrow\infty}\frac{\ln(\frac{x}{x+k})}{\frac{1}{x}}}$$

From here you should be able to use l'hopital's rule!

Answer 3

Why do you make it so complicated ? By continuity of the inverse function,

$$\frac1{\lim_{x\to\infty}\left(\dfrac x{x+k}\right)^x}=\lim_{x\to\infty}\left(1+\dfrac kx\right)^x.$$