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Why is $$L=\lim_{x\to 0} (1+x)^{1/x}=e$$

I understand that if we take log on both sides and solve we get $L=e$ but if we inspect right hand limit, then base is slightly greater than $1$ and exponent approaches $+\infty$, hence overall value must approach $+ \infty$ . If we inspect Left hand Limit, then base is slightly less than $1$ and exponent approaches $-\infty$, hence overall expression should again approach $+\infty$. Then why is limit approaching towards $e$.

Mathematics
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  • Intuitively, it depends on how fast base is moving towards $1$ while exponent approaches $\pm \infty$. In this case, both move at relatively the same speed so the result is a constant. If say the exponent move towards $\pm \infty$ at a much faster rate than the base moving towards $1$, then the result is $\pm \infty$. –  Nov 01 '17 at 02:24
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    That's exactly why $1^\infty$ is an indeterminate form: the value of $\lim_{x \to a} f(x)^{g(x)}$, if it exists, depends on the specific rate at which $f(x)$ approaches $1$ and $g(x)$ approaches $\infty$ as $x \to a$. – Alex Provost Nov 01 '17 at 02:25
  • $$\lim_{n\to 0} \left[(1+x)^{1/x}\right]\neq \left[\lim_{n\to 0}(1+x)\right]^{\lim_{n\to 0}1/x}$$ – Dave Nov 01 '17 at 02:25
  • The best way to counter your common sense is to use a calculator and evaluate the expression for $x=0.1,0.01,-0.1,-0.01$. – Paramanand Singh Nov 01 '17 at 02:42

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