Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm exercising on a book and I'm stuck at the following task: compute the $\lim\limits_{x\to\infty}\frac{a^x-1}{x}$, when $a > 1$.

As I can see, when $x\to-\infty$, the numerator tends to $-1$ and denominator tends to $-\infty$, so the whole expression tends to $0$.

But when $x\to+\infty$, the whole expression becomes indeterminate form of $\frac{\infty}{\infty}$. Well, $\lim\limits_{x\to+\infty}\frac{a^x-1}{x} = \lim\limits_{x\to+\infty}\frac{a^x}{x} - \lim\limits_{x\to+\infty}\frac{1}{x} = \lim\limits_{x\to+\infty}\frac{a^x}{x}$. Intuitively (and well proved by l'Hôpital's rule) $\frac{a^x}{x}\to+\infty$ when $x\to+\infty$. But this rule is discussed later in the book, so I'm supposed to prove this statement by means of intuitive definition of limit or so.

Since $x\to+\infty$, there must be preassigned positive $M$ conforming to $|\frac{a^x}{x}|>M$ when $x>N$, for any preassigned positive $N$. If we express $M$ in terms of $N$: $M=\frac{a^N}{N}$, we can see, that for any $N>0$ there is such $M$. So, by definition, $\lim\limits_{x\to+\infty}\frac{a^x}{x}=+\infty$.

Is my reasoning correct?

share|cite|improve this question
Your reasoning seems right, but are you sure that you are not being asked about the $\lim_{x \rightarrow 0}$? – Ron Gordon Jan 25 '13 at 5:39
Your reasoning looks backwards to me. You need to show that for every $M$, there exists an $N$ such that $x > N$ implies $\frac{a^x}{x} > M$. – user7530 Jan 25 '13 at 5:42
@rlgordonma Yes, I'm pretty sure in this. Otherwise, it would be toooooo simple:-) – Igor Gorbunov Jan 25 '13 at 5:42
Weird. It is more interesting to look at $x\to 0$. – Pedro Tamaroff Jan 25 '13 at 5:46
@user7530 Oh, that's it, we must express $N$ in terms of $M$, something like $\ln M=N\cdot\ln a - \ln N$ – Igor Gorbunov Jan 25 '13 at 6:02
up vote 2 down vote accepted

The reasoning is incomplete. It merely asserts that what you hope to be true is true.

One can come up with a more rigorous argument. It is not hard to show that $a^x$ is increasing. Let $a=1+d$. Then if $n$ is an integer $\ge 2$, by the Binomial Theorem we have $$a^n=(1+d)^n \ge 1+dn +\frac{n(n-1)}{2}\gt \frac{n(n-1)}{2}.$$ Thus if $x\ge 2$, $$\frac{a^x}{x} \gt \frac{\lfloor x\rfloor(\lfloor x\rfloor -1)}{2x}.$$
It is then quite easy to show that by choosing $x$ large enough, we can make the right-hand side above as close to $0$ as we wish.

share|cite|improve this answer
Shouldn't the latter expression be written as $\frac{a^x}{x}\ge\frac{\lfloor x\rfloor(\lfloor x\rfloor -1)}{2x}$? Then, $\frac{a^x}{x}$ really tends to $+\infty$ – Igor Gorbunov Jan 25 '13 at 6:58
@IgorGorbunov: Yes, thank you, it is $\ge$, actually $\gt$. – André Nicolas Jan 25 '13 at 7:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.