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I've tried substituting ${\frac{1}{x}}$ as $y$ and then I get $\lim_{y \to \ 0}\ (1)^{\frac{1}{y}}$ and that's infinity?

$$\lim_{x \to \infty}\ (1+ {\frac{1}{x}})^{x}$$

Martin Sleziak
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Hari
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    This is a standard limit and it is equal to $\ e$ Here's a help https://math.stackexchange.com/questions/882741/limit-of-1-x-nn-when-n-tends-to-infinity – CTSnake Apr 28 '17 at 05:51
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    Try $z(x) = \log[(1 + 1/x)^x]$ and find $\lim_{x\rightarrow\infty} z(x)$. – Michael Apr 28 '17 at 05:52
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    If you substitute $\frac{1}{x}$ as $y$, you should get $\lim_{y\to 0}(1+y)^{\frac{1}{y}}$. – GAVD Apr 28 '17 at 05:54

2 Answers2

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Hint

$$\text{If }L=\lim_{x\rightarrow a}f(x)^{g(x)}=(\rightarrow1)^{(\rightarrow\infty)}$$ then $$L=e^{\lim_{x\rightarrow a}g(x)(f(x)-1)}$$

The Dead Legend
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$\frac{d}{dx} \log x\mid_{x = 1} = \lim_{h \to 0} \frac{\log(1+h)-\log 1}{h} = \lim_{h \to 0} \log(1+h)^{1/h} = \lim_{h \to \infty} \log (1+\frac{1}{h})^h = $ $\log \lim_{h \to \infty} (1+\frac{1}{h})^h$.

But we know $\frac{d}{dx} \log x\mid_{x=1} = 1$, so it must be that $\lim_{h \to \infty} (1+\frac{1}{h})^h = e$.

mathworker21
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