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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?

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

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    $\begingroup$ 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 $+\infty$ $\endgroup$ – Igor Gorbunov Jan 25 '13 at 6:58
  • $\begingroup$ @IgorGorbunov: Yes, thank you, it is $\ge$, actually $\gt$. $\endgroup$ – André Nicolas Jan 25 '13 at 7:12

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