Calculate a limit of exponential function Calculate this limit:
$$
\lim_{x \to \infty } = \left(\frac{1}{5} + \frac{1}{5x}\right)^{\frac{x}{5}}
$$
I did this:
$$
\left(\frac{1}{5}\right)^{\frac{x}{5}}\left[\left(1+\frac{1}{x}\right)^{x}\right]^\frac{1}{5}
$$
$$
\left(\frac{1}{5}\right)^{\frac{x}{5}}\left(\frac{5}{5}\right)^\frac{1}{5}
$$
$$
\left(\frac{1}{5}\right)^{\frac{x}{5}}\left(\frac{1}{5}\right)^\frac{5}{5}
$$
$$
\lim_{x \to \infty } = \left(\frac{1}{5}\right)^\frac{x+5}{5}
$$
$$
\lim_{x \to \infty } = \left(\frac{1}{5}\right)^\infty = 0
$$
Now I checked on Wolfram Alpha and the limit is $1$
What did I do wrong? is this the right approach? is there an easier way?:)
Edit:
Can someone please show me the correct way for solving this? thanks.
Thanks
 A: The limit is indeed $0$, but your solution is wrong.
$$\lim_{x\to\infty}\left(\frac15 + \frac1{5x}\right)^{\!x/5}=\sqrt[5\,]{\lim_{x\to\infty}\left(\frac15\right)^{\!x}\lim_{x\to\infty}\left(1 + \frac1x\right)^{\!x}}=\sqrt[5\,]{0\cdot e}=0$$
And WolframAlpha confirms it: https://www.wolframalpha.com/input/?i=%281%2F5%2B1%2F%285x%29%29%5E%28x%2F5%29+as+x-%3Einfty
A: $$\lim_{x\to\infty}\left(\frac15\right)^x=0$$
$$\lim_{x\to\infty}\left(1+\frac1x\right)^x=e$$
A: The limit is $0$, and Wolfram Alpha seems to agree with this (maybe you entered the expression wrongly?)
$$\lim_{x\to\infty}\left(\frac{1}{5}\right)^{\frac{x}{5}}\left(\left(1+\frac{1}{x}\right)^x\right)^\frac{1}{5}=0\cdot e^{1/5}=0$$
by the definition of $e$, which is $e=\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x$.
A: Write
$$
\left(\frac{1+x}{5x}\right)^{x/5} = e^{\frac{x}{5}\ln\left(\frac{1+x}{5x}\right)} = e^{-\frac{x}{5}\ln\left(\frac{5x}{1+x}\right)} = e^{-\frac{x}{5}\ln\left(5-\frac{5}{1+x}\right)}
$$
Observing that $\ln\left(5-\frac{5}{1+x}\right)\xrightarrow[x\to\infty]{}\ln 5$, can you conclude by composing limits?
