calculus - 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

Answer

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