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In another answer I saw, there is this expression

$$\lim _{ x\rightarrow \infty }{ x{ \left( 1-\frac { 1 }{ x } \right) }^{ x } } =\lim _{ x\rightarrow \infty }{ \frac { x }{ e } } =\infty$$

Can anyone show me how to get to this result? There's a multiplication by $x$, a division by $x$, and an exponent of $x$, so it seems like a good example to figure out how all of this interplays with each other when raising $x$ to infinity.


marked as duplicate by Dietrich Burde, Mark Fantini, Chappers, Jonas Meyer, Shuchang Apr 2 '15 at 3:59

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  • $\begingroup$ Do you know that $\lim_{x\to\infty}\left(1+\frac yx\right)^x=e^y$? $\endgroup$ – Regret Apr 1 '15 at 18:56
  • $\begingroup$ It's an application of $$\lim_{x \to \infty}\big(f(x)g(x)\big) =\left(\lim_{x \to \infty}f(x)\right)\left(\lim_{x \to \infty}g(x)\right)$$ $\endgroup$ – Arthur Apr 1 '15 at 18:56
  • $\begingroup$ @Regret I do now! Why is that limit the case? $\endgroup$ – user51819 Apr 1 '15 at 18:56
  • $\begingroup$ @Arthur So I am curious, what would happen if the left side were -infinity and the right side were +infinity? Would it just net out to -infinity? $\endgroup$ – user51819 Apr 1 '15 at 18:58
  • $\begingroup$ Yes, it would. The only problematic case is $\pm \infty\cdot 0$. $\endgroup$ – Arthur Apr 1 '15 at 18:59

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