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

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.