Is it true that f(x) decreases faster than g(x).

Question

Suppose we have two functions

$f(x) = a^x$

$g(x) = b^x$

Now suppose that $0 < a,b < 1$ and that $a > b$, is it true that $f(x)$ will decrease faster than $g(x)$ as $x \to \infty$?

I attempted to show that it is true by using the following reasoning (Apologies for the rather crude attempt):

Suppose that $a > b$ and that $a, b > 1$. I believe it's rather intuitive to say that $f$ will grow exponentially faster than $g$ as $x \to \infty$. We can also observe that as $x$ decreases (but does not decrease below $0$), $f(x)$ decreases at a faster rate than $g(x)$.

My question is, suppose $x$ moves from $0$ to $-\infty$, is it true that $f$ decreases faster than $g$? (Still under the condition that $a > b$ and that $a, b > 1$)

I think if the answer to the question above is true, then it can be implied that the answer to the initial question is true. Is this way of thinking correct?

Answer 1

$$f'(x) = a^x \log a$$ $$g'(x) = b^x \log b$$ So

$$h(x) =\frac{f'(x)}{g'(x)} = c \, (\frac ab)^x $$ where $c > 0$ since $\log a > \log b$. Furthermore, $\frac ab > 1$.

And so $$\lim_{x \rightarrow -\infty} h(x) =0$$

Which means that under the condition that $a>b$ and that $a,b>1$, $g(x)$ decreases faster than $f(x)$ as $x$ gets more negative.

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We can use the same technique to answer the initial question. Again, let

$$h(x) =\frac{f'(x)}{g'(x)} = c \, (\frac ab)^x $$ where $c > 0$ since $\log a > \log b$. Furthermore, $\frac ab > 1$.

And so $$\lim_{x \rightarrow \infty} h(x) = \infty$$

Which means that under the condition that $0<a,b<1$ and $a>b$, $f(x)$ decreases faster than $f(x)$ as $x$ increases to infinity.