My approach so far:

Using this I'm expressing the above limit as $((1+\frac{1}{\frac{n}{x_n}})^\frac{n}{x_n})^{x_n}$ and then using the property (?) that if $x_n \rightarrow x$ and $a_n \rightarrow a$, then $x^{a_n}_n \rightarrow x^{a}$. But I'm proving this last property by taking log on both sides (suppose $a,x > 0$). My question is isn't that somehow using the property I'm required to prove and hence is this valid?

  • $\begingroup$ The link you provide seems to have an answer (the second most upvoted one citing Königsberger). $\endgroup$ – AddSup Oct 6 '18 at 11:56
  • $\begingroup$ Another one: math.stackexchange.com/q/374747/42969. $\endgroup$ – Martin R Oct 6 '18 at 12:03

Use the fact that $(1+\frac {x+\epsilon} n)^{n} \to e^{x+\epsilon}$ and $(1+\frac {x-\epsilon} n)^{n} \to e^{x-\epsilon}$ and use squeeze theorem.

  • $\begingroup$ Thank you so much! That's so simple and useful. Here's what I've understood of the solution: So suppose $x>0$, then $(1+\frac {x}{n})^{n}$ is increasing in $x$ for any given $n$. That gives me $e^{x-\epsilon} \leq \overline{lim} (1+\frac {x_n}{n})^{n} \leq e^{x+\epsilon}$ for all $\epsilon$. Same holds for $\underline{lim}$. Then we take $lim\; \epsilon \rightarrow 0$. $\endgroup$ – Canine360 Oct 6 '18 at 12:15
  • 1
    $\begingroup$ @Canine360 Yes, that is exactly what the argument is. $\endgroup$ – Kavi Rama Murthy Oct 6 '18 at 12:16

Assume $x_n \not =0.$

$y:=x_n/n \not =0$, and $\lim_{n \rightarrow \infty} y_n=0.$


$\exp(\log (1+y_n)^n)$.


$n\log(1+y_n)= $

$x_n \dfrac{\log (1+y_n) -\log 1}{y_n}=$

$x_n \log '(t_n) = 1/t_n$, where $t_n \in (1, 1+y_n)$.


$ \lim_{ n \rightarrow \infty} y_n =0$ implies

$ \lim_{n \rightarrow \infty} t_n =1.$


$\lim_{n \rightarrow \infty}(x_n)(1/t_n)=$

$\lim_{n \rightarrow \infty}x_n \lim_{n \rightarrow \infty}(1/t_n)= x.$

Finally, using the continuity of the exponential function:

$\lim_{n \rightarrow \infty}(1+x_n/n)^n= e^x.$


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