$\lim_{x\to +\infty}2^{-x}*(2+\frac{1}{x})^x$

Question

I know that this limit is $\sqrt e$, and I can arrive at this, but I could also say

$\lim_{x\to +\infty}2^{-x}*(2+\frac{1}{x})^x = \lim_{x\to \infty}2^{-x}*(2)^x = 1$

What is wrong with this? Maybe the fact that I am using asymptotic in a sum which is raised at x?

Answer 1

HINT

As noticed your way is wrong, we can proceed as follow

$$2^{-x}\left(2+\frac{1}{x}\right)^x = \left(1+\frac{1}{2x}\right)^x=e^{\frac12\log \frac{\left(1+\frac{1}{2x}\right)}{\frac{1}{2x}}}$$

Answer 2

Notice that $$2^{-x}(2+\dfrac{1}{x})^x=(1+{1\over 2x})^x=\sqrt{(1+{1\over 2x})^{2x}}$$

Answer 3

You assumed that $(2+\frac{1}{x})^x=2^x$ which is wrong. If that was correct then $(1+\frac{1}{x})^x=1$ which we know does not hold because $(1+\frac{1}{x})^x=e$

What I am trying to say is that you are not allowed to substitute the limits with the limit value or to do such tricks. That's why there are some rules on limits that might seem obvious but we must conform to them and only them!