Thursday 27 March 2014

calculus - Prove the limit is $sqrt{e}$.





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.


Answer



Hint: $$\frac{a^k}{b^k}=\left(\frac ab\right)^k,$$ and in general, $$\lim_{k\to\infty}\left(1+\frac xk\right)^k=e^x.$$


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