Solving Indeterminate Limits with De L'Hopital's Rule

[Felafel]

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..

 

[Millennial]

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.

 

[Felafel]

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]

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)\ .

 

[Felafel]

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.

 

[Dick]

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).

 

[Felafel]

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]

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.