# Show that $x_n \rightarrow x \Rightarrow (1+\frac{x_n}{n})^n \rightarrow e^x$ [duplicate]

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?

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

• 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 $lim\; \epsilon \rightarrow 0$. – Canine360 Oct 6 '18 at 12:15
• @Canine360 Yes, that is exactly what the argument is. – Kavi Rama Murthy Oct 6 '18 at 12:16

Assume $$x_n \not =0.$$

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

$$(1+y_n)^n=$$

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

Consider:

$$n\log(1+y_n)=$$

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

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

Note:

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

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

Hence

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

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

Finally, using the continuity of the exponential function:

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