# Another difficult limit

## Homework Statement

here it is:
the teacher suggeste to solve it with de l'hopital, however it is not necessary

$\displaystyle \lim x \to +\infty\ (\frac{a^x -1}{x(a-1)})^\frac{1}{x}$
with $a>0$ , $a≠1$

## The Attempt at a Solution

the indeterminate formula is $(\frac{\infty}{\infty})^0$
but i thought of changing the variable $x \to \infty$ to $z \to 0$
such that i have the known:
$\frac{(a^{\frac{1}{t}}-1)}{\frac{1}{t}} = log(a)$
so i obtain:
$(\frac{log(a)}{a-1})^{+\infty}$
but i don't know how to go on..

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Why don't you simply use the fact that the limiting operator respects division when the limits on the numerator and the denominator are defined? It simplifies the problem to a great extent.

Why don't you simply use the fact that the limiting operator respects division when the limits on the numerator and the denominator are defined? It simplifies the problem to a great extent.
ok, then it should be:

$\displaystyle \lim z \to 0 (\frac{a^{(1/z)-1}}{1/z})^z * \displaystyle \lim z \to 0 (\frac{1}{a-1})^z$
$=(log(a))^0*(\frac{1}{a-1})^0$
=1

is it correct?

SammyS
Staff Emeritus
Homework Helper
Gold Member
Your post #3 result is incorrect.

## Homework Statement

here it is:
the teacher suggested to solve it with de l'hopital, however it is not necessary

$\displaystyle \lim x \to +\infty\ (\frac{a^x -1}{x(a-1)})^\frac{1}{x}$
with $a>0$ , $a≠1$

## The Attempt at a Solution

the indeterminate formula is $(\frac{\infty}{\infty})^0$
but i thought of changing the variable $x \to \infty$ to $z \to 0$
such that i have the known:
$\frac{(a^{\frac{1}{t}}-1)}{\frac{1}{t}} = log(a)$
so i obtain:
$(\frac{log(a)}{a-1})^{+\infty}$
but i don't know how to go on..
If $\displaystyle \lim \, \ln(f(x))=L\,,\ \text{ then }\ \lim\, f(x)=e^L\ .$

So, look at $\displaystyle \lim_{x \to +\infty\ }\ \ln\left(\left(\frac{a^x -1}{x(a-1)}\right)^{1/x}\right)\ .$

I've found a possible solution, but I'm a bit dubious, could you please check it?
1st situation: $0<a<1$

$\displaystyle \lim_{x \to +\infty} (\frac{a^x-1}{x(a-1)})^{\frac{1}{x}}=\displaystyle \lim_{x \to +\infty} e^{\frac{1}{x}log(\frac{a^x-1}{x(a-1)})}$

If $\lim log(f(x)) = L$ then $\lim f(x) = e^L$

$=\frac{1}{+\infty} log(\frac{-1}{-\infty})(=\frac{-\infty}{+\infty})$

With de l'Hopital

$\stackrel{\text{H}}{=} \frac{x(a-1)}{a^x-1} \frac{x a^x log(a) (a-1) - (a-1) (a^x-1)}{x^2 (a-1)^2} = \frac{x a^x log(a) - (a^x-1)}{(a^x-1)x}$

= $\frac{a^xloga}{a^x-1}-\frac{1}{x}$

$= \frac{0}{-1}-0=0$

$e^0 = 1$

Situation 2: $a>1$

$\displaystyle \lim_{x \to +\infty} e^{\frac{1}{x}log(\frac{a^x-1}{x(a-1)})}=$
$=e^{ 0* log (\frac{\infty}{\infty})}$

I use de l'Hopital within the log (am I allowed to do this?)

$\displaystyle \lim_{x \to +\infty} \frac{1}{x}log(\frac{a^x loga}{a-1})$ = $\frac{\infty}{\infty}$

de l'Hospital again

$\displaystyle \lim_{x \to +\infty} \frac{a-1}{a^x loga} \frac{a^xlog^2a+0-0}{(a-1)^2} = \frac{loga}{a-1}$

$=e^{\frac{loga}{a-1}}$

I didn't want to use the known limit = $\frac{a^x-1}{x}=loga$ because x had to go to 0 to apply that, while here it doesn't.

Last edited:
Dick
Homework Helper
I've found a possible solution, but I'm a bit dubious, could you please check it?
1st situation: $0<a<1$

$\displaystyle \lim_{x \to +\infty} (\frac{a^x-1}{x(a-1)})^{\frac{1}{x}}=\displaystyle \lim_{x \to +\infty} e^{\frac{1}{x}log(\frac{a^x-1}{x(a-1)})}$

If $\lim log(f(x)) = L$ then $\lim f(x) = e^L$

$=\frac{1}{+\infty} log(\frac{-1}{-\infty})(=\frac{-\infty}{+\infty})$

With de l'Hopital

$\stackrel{\text{H}}{=} \frac{x(a-1)}{a^x-1} \frac{x a^x log(a) (a-1) - (a-1) (a^x-1)}{x^2 (a-1)^2} = \frac{x a^x log(a) - (a^x-1)}{(a^x-1)x}$

= $\frac{a^xloga}{a^x-1}-\frac{1}{x}$

$= \frac{0}{-1}-0=0$

$e^0 = 1$

Situation 2: $a>1$

$\displaystyle \lim_{x \to +\infty} e^{\frac{1}{x}log(\frac{a^x-1}{x(a-1)})}=$
$=e^{ 0* log (\frac{\infty}{\infty})}$

I use de l'Hopital within the log (am I allowed to do this?)

$\displaystyle \lim_{x \to +\infty} \frac{1}{x}log(\frac{a^x loga}{a-1})$ = $\frac{\infty}{\infty}$

de l'Hospital again

$\displaystyle \lim_{x \to +\infty} \frac{a-1}{a^x loga} \frac{a^xlog^2a+0-0}{(a-1)^2} = \frac{loga}{a-1}$

$=e^{\frac{loga}{a-1}}$

I didn't want to use the known limit = $\frac{a^x-1}{x}=loga$ because x had to go to 0 to apply that, while here it doesn't.
No, don't try the "l'Hopital within the log". You are making life a lot harder by not using rules of logs. Split your log up into log(a^x-1)-log(x)-log(a-1).

oh, allright, i didn't think about it. so now it should be:

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

using de l'Hopital i get:

$\frac{1}{a^x-1}log(a)-\frac{1}{x}-\frac{1}{a-1}$ diving both numerator and denominator of the first member it becomes

$\frac{log(a)}{1-\frac{1}{a^x}}-\frac{1}{x}-\frac{1}{a-1}$=$log(a)-\frac{1}{a-1}$

=$\frac{e^{log(a)}}{e^{\frac{1}{a-1}}}$

Is everything ok now?
at the beginning was it good to divide the limit in two different cases?
Thanks a lot again

Dick
Homework Helper
oh, allright, i didn't think about it. so now it should be:

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

using de l'Hopital i get:

$\frac{1}{a^x-1}log(a)-\frac{1}{x}-\frac{1}{a-1}$ diving both numerator and denominator of the first member it becomes

$\frac{log(a)}{1-\frac{1}{a^x}}-\frac{1}{x}-\frac{1}{a-1}$=$log(a)-\frac{1}{a-1}$

=$\frac{e^{log(a)}}{e^{\frac{1}{a-1}}}$

Is everything ok now?
at the beginning was it good to divide the limit in two different cases?
Thanks a lot again
Yes, it's a very good idea to divide it into different cases. a^x behaves differently in the two cases. You are getting there. The derivative of log(a-1) is zero isn't it? It's a constant! And you forgot a factor of a^x in the first term, but the next line is correct, so I guess that's a typo.