Question

Suppose that f (x) is o(g(x)). Does it follow that $2^{f\left(x\right)}$is o( $2^{g\left(x\right)}$).

Short answer

The answer is yes.

Step-by-step solution

Step 1:

Our functions are a(x)=$2^{f\left(x\right)}$ and b(x)= $2^{g\left(x\right)}$, and we also know that $\underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}=0$

Let our old functions f(x) and g(x) be :

$$\begin{gathered} f\left(x\right)=x^2 \\ g\left(x\right)=x^3 \end{gathered}$$

Then, $\underset{x\to \mathrm{\infty }}{lim} \frac{f\left(x\right)}{g\left(x\right)}=0$

Step 2:

Determining the limit ratios of the new functions:

$\begin{array}{r}\underset{x\to \mathrm{\infty }}{lim} \frac{a\left(x\right)}{b\left(x\right)}=\underset{x\to \mathrm{\infty }}{lim} \frac{2^{x^2}}{2^{x^2}}=\underset{x\to \mathrm{\infty }}{lim} \left(2^{x^2}-2^{x^3}\right)\\ \Rightarrow \underset{x\to \mathrm{\infty }}{lim} \frac{1}{2^{x^3}-2^{x^2}}\end{array}$

Since $x^3>x^2$

$\underset{x\to \mathrm{\infty }}{lim} \frac{a\left(x\right)}{b\left(x\right)}=\underset{x\to \mathrm{\infty }}{lim} \frac{1}{2^{x^3}-2^{x^2}}=\frac{1}{\mathrm{\infty }}=0$

We see that a(x) is o(b(x)).