To evaluate $\lim_{n \to \infty} \left(1+\frac{1}{2n}\right)^{n+5}$ To evaluate 
$$\lim_{n \to \infty} \left(1+\frac{1}{2n}\right)^{n+5}$$
I write it as 
$$\lim_{n \to \infty} \left(1+\frac{1}{2n}\right)^{n}\left(1+\frac{1}{2n}\right)^5.$$
My question is how to evaluate
$$\lim_{n \to \infty} \left(1+\frac{1}{2n}\right)^{n}.$$
 A: $$\lim_n\left(1+\frac{1}{2n}\right)^{n}=\left[\lim_n\left(1+\frac{1}{2n}\right)^{2n}\right]^{1/2}= \sqrt{e}$$
A: $(1+\frac{1}{2n})^{n+5}=\left(1+\dfrac{\frac{1}{2}}{n}\right)^n(1+\frac{1}{2n})^5\to e^{\frac{1}{2}}$
A: Just another way $$A_n=(1+\frac{1}{2n})^{n+5}$$ Taking logarithms $$\log(A_n)=(n+5)\log(1+\frac{1}{2n})$$ Now, using that, for small $x$, $\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)$, replace $x$ by $\frac{1}{2n}$ to get $$\log(A_n)=\frac{1}{2}+\frac{19}{8 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and also how it is approached.
A: $$\lim_{n \to \infty} (1+\frac{1}{2n})^{n}(1+\frac{1}{2n})^5 =\lim_{n \to \infty} (1+\frac{1}{2n})^{n} = \lim_{n \to \infty} \left(1+\frac{1 \over 2}{n} \right)^{n}  = e^{1 \over 2} = \sqrt{e}$$
A: $$\lim_{n\to\infty}\left(1+\frac{1}{2n}\right)^{n+5}=\lim_{n\to\infty}\left(1+\frac{1}{2n}\right)^{n}\cdot\lim_{n\to\infty}\left(1+\frac{1}{2n}\right)^{5}$$
$$=\lim_{n\to\infty}\left(1+\frac{1}{2n}\right)^{2n\cdot\frac{1}{2}}\cdot\lim_{n\to\infty}\left(1+\frac{1}{2n}\right)^{5}=e^{\frac{1}{2}}\cdot 1^5=e^{\frac{1}{2}},$$
becouse $\lim_{n\to\infty}{\left(1+\frac{1}{n}\right)}^n=e, \text{and} $$\lim_{n\to\infty}\frac{1}{2n}=0$
