How do you show $$\lim\limits_{k \rightarrow \infty} \frac{\left(2+\frac{1}{k}\right)^k}{2^k}=\sqrt{e}$$
I know that $$\lim\limits_{k \to \infty} \left(1+\frac{1}{k}\right)^k=e$$ but I don't know how to apply this.
$\endgroup$ 02 Answers
$\begingroup$Hint: $$\frac{a^k}{b^k}=\left(\frac ab\right)^k,$$ and in general, $$\lim_{k\to\infty}\left(1+\frac xk\right)^k=e^x.$$
$\endgroup$ 1 $\begingroup$Hint:$$\frac{\Bigl(2+\dfrac{1}{k}\Bigr)^k}{2^k}=\Bigl(1+\dfrac{1}{2k}\Bigr)^k=\sqrt{\Bigl(1+\dfrac{1}{2k}\Bigr)^{2k}}.$$
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