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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

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  • $\begingroup$ WOlfram alpha gives limit as $0$. However, if $x \rightarrow 0$, then the limits is $1$. $\endgroup$ – voldemort Dec 20 '14 at 16:00
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    $\begingroup$ The first error was $(1+\frac{1}{x})^x\neq \frac{5}{5}$. $\endgroup$ – vadim123 Dec 20 '14 at 16:00
  • $\begingroup$ but it gives you $1^{\frac{1}{5}}$ so $\frac{5}{5}^{\frac{1}{5}}$ equals $\frac{1}{5}^{\frac{5}{5}}$ $\endgroup$ – FigureItOut Dec 20 '14 at 16:06
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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

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$$\lim_{x\to\infty}\left(\frac15\right)^x=0$$

$$\lim_{x\to\infty}\left(1+\frac1x\right)^x=e$$

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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 $e=\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x$.

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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?

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