What is the value of $\lim_{x\to \infty} \frac{a^x-1}{x}$ for a > 1

Question

I tried the following: $$ \lim _{x\rightarrow \infty }\dfrac {a^{x}\left( 1-\dfrac {1}{a^{x}}\right) }{x} \\ = \lim _{x\rightarrow \infty }\dfrac {a^{x}}{x}•\lim _{x\rightarrow \infty }\left( 1-\dfrac {1}{a^{x}}\right)\\ $$ I did realize that $a^x > x$ $ \forall$ $ x\epsilon R$.
But this doesn't suffice to say that the first limit in the second expression tends to infinity since we still haven't proven $a^x$ is an infinity of higher order than $x$.

Answer 1

Note that for $x\to + \infty$

$$\frac{a^x-1}{x}=\frac{a^x-1}{a^x}\frac{a^x}{x}\to (1 \cdot +\infty)\to +\infty$$

indeed

• $\frac{a^x-1}{a^x}\to 1$

and since eventually $a^x\ge x^2$

• $\frac{a^x}{x}\ge\frac{x^2}{x}= x\to \infty$

For $x\to - \infty$ let $y=-x\to +\infty$

$$\lim _{x\to -\infty } \frac {a^x-1}{x}=\lim _{y\to \infty } \frac {a^{-y}-1}{-y}=\lim _{y\to \infty } \frac {a^y-1}{ya^y}=\lim _{y\to \infty } \frac1y \frac {a^y-1}{a^y}=0$$

Answer 2

Use equivalents:

if $a>1$, $a^x-1\sim_{+\infty} a^x=\mathrm e^{x\ln a}$, so $$ \frac{a^x-1}{x}\sim_{+\infty}\frac{e^{x\ln a}}x\to +\infty\quad\text{since }\ln a>0.$$

At $-\infty$, $\;a^x-1\sim -1$, so $$\frac{a^x-1}{x}\sim_{-\infty}-\frac1x\to 0.$$