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Evaluate $\lim_{x\to\infty}\left(1+\frac 4x\right)^{\frac x8}$

I think the end result is $1^\infty$, so the answer is undefined?

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    $\begingroup$ Hint: are you familiar with the limit definition of $e^x$? $\endgroup$ – abiessu May 22 '15 at 21:27
  • $\begingroup$ $1^{\infty}$ is an indeterminate form. What this means is precisely that knowing $\lim f(x)=1$ and $\lim g(x)=\infty$, you cannot say anything about $\lim \left(f(x)^{g(x)}\right)$ just using this fact alone. Don't confuse 'indeterminate' with 'undefined'. $\endgroup$ – user26486 May 22 '15 at 22:37
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$$\lim_{x\to\infty}\left(1+\frac 4x\right)^{\frac x8}=$$ $$\lim_{x\to\infty} \exp\left(\frac{1}{8}x\ln\left(1+\frac{4}{x}\right)\right)=$$ $$\exp\left(\lim_{x\to\infty} \frac{\ln\left(1+\frac{4}{x}\right)}{\frac{8}{x}}\right)=$$ $$\exp\left(\lim_{x\to\infty}\frac{x}{2(x+4)}\right)=$$ $$\exp\left(\lim_{x\to\infty}\frac{1}{2}\right)=$$ $$e^{\frac{1}{2}}=\sqrt{e}$$

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One of the definitions of $e^y$ is $\lim_{x \to \infty} \left(1+\dfrac{y}x\right)^x$. I trust you can conclude from this.

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  • $\begingroup$ Oh, you were faster, I thought the same. :) $\endgroup$ – Atvin May 22 '15 at 21:25
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Hint: You can use the fact, that $e^x$ is the $\lim_{n \to \infty} \left(1+\dfrac{x}n\right)^n$

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What you can also do is to write out the following, $$ \lim_{x \rightarrow \infty} (1+ \frac{4}{x})^\frac{x}{8} = \lim_{x \rightarrow \infty} e^{\ln \left( (1+ \frac{4}{x})^\frac{x}{8} \right)} = \lim_{x \rightarrow \infty} e^{\frac{x}{8} ln( 1 + \frac{4}{x})} = e^{ \lim_{x \rightarrow \infty} \frac{x}{8} ln( 1 + \frac{4}{x})} $$ because $e^x$ is continuous. Now all we need is to determine the following limit and plug it into $e^x$, $$ \lim_{x \rightarrow \infty} \frac{x}{8} \ln (1 + 4/x) $$ From L'hopitals rule, the limit is $\frac{1}{2}$. So the answer we are looking for is $\sqrt{e}$.

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