Find the value of $\lim_{n \rightarrow \infty} \sqrt{1+\left(\frac1{2n}\right)^n}$

Find the limit of the sequence as it approches $\infty$ $$\sqrt{1+\left(\frac1{2n}\right)^n}$$

I made a table of the values of the sequence and the values approach 1, so why is the limit $e^{1/4}$?

I know that if the answer is $e^{1/4}$ I must have to take the $\ln$ of the sequence but how and where do I do that with the square root?

I did some work getting the sequence into an indeterminate form and trying to use L'Hospitals but I'm not sure if it's right and then where to go from there. Here is the work I've done

$$\sqrt{1+\left(\frac1{2n}\right)^n} = \frac1{2n} \ln \left(1+\frac1{2n}\right) = \lim_{x\to\infty} \frac1{2n} \ln \left(1+\frac1{2n}\right) \\ = \frac 1{1+\frac1{2n}}\cdot-\frac1{2n^2}\div-\frac1{2n^2}$$

Thank you

• You've written two different sequences in your question. The first one you mention is $\sqrt{1 + (1/2n)^n}$, and the second is $(1 + (1/2n))^{n/2}$. – Christopher A. Wong Mar 26 '13 at 20:29
• I think you have messed up with brackets, because the original sequence has the limit of 1 indeed. – rtybase Mar 26 '13 at 20:29
• @user68626 I fixed the formatting. In the future, look at meta.math.stackexchange.com/questions/5020/…. – PyRulez Mar 26 '13 at 20:40
• The problem is that your problem is wrong - you want $$\lim_{n\to\infty}\sqrt{\left(1+\frac{1}{2n}\right)^n}$$ The limit, as you wrote it, is $1$, the limit I've written is $e^{1/4}$. – Thomas Andrews Mar 26 '13 at 20:55

Hint: $$\left(1+\frac{1}{2n}\right)^n=e^{n\log\left(1+\frac1{2n}\right)}\sim e^{n\times \frac 1{2n}}=e^{\frac12}$$ So what's the desired limit?
We have $\lim_{n\to\infty} (1+\frac{x}{n})^n = e^x$. Hence $\lim_{n \to \infty} (1+\frac{x}{2n})^{2n} = e^x$, then taking square roots (noting that the square root is continuous on $[0,\infty)$), we have $\lim_{n \to \infty} \sqrt{(1+\frac{x}{2n})^{2n}} = \lim_n (1+\frac{x}{2n})^{n} = \sqrt{e^x} = e^\frac{x}{2}$, and finally $\lim_{n\to\infty} \sqrt{(1+\frac{x}{2n})^{n}} = e^\frac{x}{4}$.
Setting $x=1$ gives the desired result.