lower bounding exponential functions

Question

I am struggling to understand constant lower bounds for some special form of exponential functions. I am aware of the following result: $\lim_{n\to \infty}(1-\frac{x}{n})^n = e^{-x}$.

But how do I get constant lower bounds like

$$\left(1-\frac{2^x}{10n}\right)^{\Large\frac{n}{2^x}} \ge 2^{-\large\frac{1}{10}}$$

or

$$\left(1-\frac{2^x}{10n}\right)^{\Large\frac{n}{2^{(x-1)}}} \ge 4^{-\large\frac{1}{5}}$$

Answer 1

I have found the following estimation:

$4^{-x} \le 1-x \le 2^{-x}$

where the first inequality hold for $0 \lt x \le \frac{1}{2}$ and the second for $0 \lt x \lt 1$.

With some basic arithmetic rules we can get the bounds asked for in the question.

(For others it might also be helpful if someone could provide a proof)

Answer 2

You can find the minimum value of the function that is on the left in your expression

$$\left(1-\frac{2^x}{10n}\right)^{\Large\frac{n}{2^x}} \ge 2^{-\large\frac{1}{10}}$$

Remember that to find the minimum value only you have that to find the derivate and to equal it to 0