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What is the result of :

\begin{equation*} \lim_{y \rightarrow \infty}( \lim_{x \rightarrow \frac{1}{y}} 1+x)^y \end{equation*}

Omar Elawady
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1 Answers1

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$$\lim \limits_{y \to \infty} \left(\lim \limits_{x \to \frac{1}{y}} (1 + x)\right)^y = \lim \limits_{y \to \infty}\left(1 + \frac{1}{y}\right)^y = e$$

Dominik
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  • And if I replaced 1 over y by 0, Would it be the same ? – Omar Elawady Dec 05 '16 at 13:29
  • No, then the limit would be $1$. – Dominik Dec 05 '16 at 13:29
  • But if I evaluated the inner limit taking into account that y approaches infinity and as a result x approaches zero, the result of it would be exactly one because limits doesn't produce infinitesimal values. – Omar Elawady Dec 05 '16 at 13:32
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    No, you need to calculate the limits in the right order. You can't just plug in the limiting values in any order. For a simpler example, note that $1 = \sum \limits_{i = 1}^n \frac{1}{n}$ does not converge to zero. – Dominik Dec 05 '16 at 13:35
  • And I guess that right order implies the evaluation of the innermost limit first. And I don't understand the link between plugging the limiting value and the result of the summation. – Omar Elawady Dec 05 '16 at 13:43
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    You try to calculate $\lim \limits_{y \to \infty} a(y)$, where the function $a(y)$ is given by $a(y) = \left(\lim \limits_{x \to \frac{1}{y}}(1 + x)\right)^y$. From the standard rules for limits we can infer $a(y) = \left(1 + \frac{1}{y}\right)^y$, which yields the result in my post. As for my example, one could argue that as $n$ approaches infinity, the value of $\frac{1}{n}$ goes to zero and we get the sum over zero, which is zero. But this is clearly not the case. – Dominik Dec 05 '16 at 13:46