Prove that $\lim_{n \to \infty}\bigg(1+\frac{\alpha}{f(n)}\bigg)^{f(n)}=e^\alpha$

Question

how to prove that

$$\lim_{n \to \infty}\bigg(1+\frac{\alpha}{f(n)}\bigg)^{f(n)}=e^\alpha$$

where $f(n)$ is any function, such $$\lim_{n \to \infty}f(n)=\infty$$

Answer 1

It's immediate from:

$$\lim_{n \to \infty}\bigg(1+\frac{\alpha}{f(n)}\bigg)^{f(n)}=\lim_{f(n) \to \infty}\bigg(1+\frac{\alpha}{f(n)}\bigg)^{f(n)}=e^\alpha$$

As an alternative by taylor expansion:

$$\bigg(1+\frac{\alpha}{f(n)}\bigg)^{f(n)}=e^\alpha=e^{f(n)\log \bigg(1+\frac{\alpha}{f(n)}\bigg)}=e^{f(n)\bigg(\frac{\alpha}{f(n)}+o\left(\frac{1}{f(n)}\right)\bigg)}=e^{\alpha+o(1)}\to e^\alpha$$

Answer 2

If $$\lim_{n \to \infty}f(n)=\infty$$ then it follows immediately that $$\lim_{n \to \infty}\bigg(1+\frac{\alpha}{f(n)}\bigg)^{f(n)}=\lim_{n \to \infty}\bigg(1+\frac{\alpha}{n}\bigg)^{n}=e^\alpha.$$ You should probably think about why that is...

Answer 3

Note that $$ \lim_{n \to \infty}\bigg(1+\frac{\alpha}{f(n)}\bigg)^{f(n)}=\lim_{n \to \infty} \left( \bigg(1+\frac{1}{\frac{f(n)}{\alpha}}\bigg)^{\frac{f(n)}{\alpha}} \right)^{\alpha}. $$ So, if $n \to \infty$ then $\frac{f(n)}{\alpha} \to \infty$ and the inner part goes to $e$.