Hi ! I'm trying to find the limit of f(x) = x.ln(1-(1/x)) when x -> +infinity

[I_love_everyone]

Hi everyone and thank you for taking the time to read this.

[math]f(x) = x.ln(1-1/x)[/math]
Pretty much everything i said in the title. I've found that Df = ] -infinty ; 0[ U ] 1 ; +infinity [. So I know that finding such a limit is possible, but I struggle to find a form of f(x) that gets rid of the ambiguous product.

I've tried switching variables : X = 1/x, so that lim(x -> + infinity) x = lim(X -> 0) X.
But that does not really help since lim(X -> 0) (1/X).ln(1-X) is also ambiguous.

I've also tried messing around with the expression itself : [math]f(x) = x.ln((x-1)/x)=x(ln(x-1) - ln(x)) = x.ln(x-1) - x.ln(x)[/math]But it doesn't really get me anywhere either for lim(x -> + infinity).

I actually know for fact that the answer is -1. But how do you get to that result ?

Does anyone have a bright idea ? I thank everyone who will take the time to understand and respond to this.

PS : I'm not english. I can write decently, but I'm not very familiar with math vocabulary in english. I apologize if some parts of my question are not explained in a rigorous manner.

 

[BigBeachBanana]

Hi everyone and thank you for taking the time to read this.

[math]f(x) = x.ln(1-1/x)[/math]
Pretty much everything i said in the title. I've found that Df = ] -infinty ; 0[ U ] 1 ; +infinity [. So I know that finding such a limit is possible, but I struggle to find a form of f(x) that gets rid of the ambiguous product.

I've tried switching variables : X = 1/x, so that lim(x -> + infinity) x = lim(X -> 0) X.
But that does not really help since lim(X -> 0) (1/X).ln(1-X) is also ambiguous.

I've also tried messing around with the expression itself : [math]f(x) = x.ln((x-1)/x)=x(ln(x-1) - ln(x)) = x.ln(x-1) - x.ln(x)[/math]But it doesn't really get me anywhere either for lim(x -> + infinity).

I actually know for fact that the answer is -1. But how do you get to that result ?

Does anyone have a bright idea ? I thank everyone who will take the time to understand and respond to this.

PS : I'm not english. I can write decently, but I'm not very familiar with math vocabulary in english. I apologize if some parts of my question are not explained in a rigorous manner.

Have you studied L'hopital rule? If so, rewrite the limit as follow:
[math]\lim_{x \to \infty}x\ln\left(1-\frac{1}{x}\right) =\lim_{x \to \infty}\frac{\ln(1-\frac{1}{x})}{1/x}[/math]Now, notice that the limit has an indeterminant form of [imath]\frac{0}{0}[/imath], and L'hopital's rule is applicable.

 

[I_love_everyone]

Hi !
Thank you for taking the time to respond !
I have never heard of this rule, but i've google it, and it seems to apply to my problem very well. I can't really work on it right now, but i'll definitely get back to you in the next couple of days.

Thanks again.

 

[pka]

Hi everyone and thank you for taking the time to read this.
[math]f(x) = x.ln(1-1/x)[/math]

Hint: [imath]\mathop {\lim }\limits_{x \to \infty } x\log \left( {1 - \frac{1}{x}} \right)=\mathop {\lim }\limits_{x \to \infty } \log \left( {1 - \frac{1}{x}} \right)^x[/imath]
Then recall that [imath]\Large\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{{ax + b}}} \right)^{cx}} = {e^{ac}}[/imath]

 

[pka]

Hint: [imath]\mathop {\lim }\limits_{x \to \infty } x\log \left( {1 - \frac{1}{x}} \right)=\mathop {\lim }\limits_{x \to \infty } \log \left( {1 - \frac{1}{x}} \right)^x[/imath]
Then recall that [imath]\Large\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{{ax + b}}} \right)^{cx}} = {e^{ac}}[/imath]

CORRECTION [imath]\Large\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{{ax + b}}} \right)^{cx}} = {e^{\frac{c}{a}}}[/imath]

 

[lookagain]

Also, \(\displaystyle \ ln(1 + x) \ = \ \dfrac11x^1 - \dfrac12x^2 + \dfrac13x^3 - \dfrac14x^4 + \ ... \ \ \ \ \ for \ \ -1 < x \le 1\)

Instead, what you are taking the limit of as x approaches +oo is \(\displaystyle \ x[ln(1 - \tfrac1x)] \ = \)

\(\displaystyle x\bigg[\bigg( \dfrac{-1}{x}\bigg)^1 \ - \ \dfrac{1}{2}\bigg( \dfrac{-1}{x}\bigg)^2 \ + \ \dfrac{1}{3}\bigg( \dfrac{-1}{x}\bigg)^3 \ - \ ... \ \ \bigg ] \ = \ \) ?

 

[I_love_everyone]

Hi everyone and thank you so much for all the help you've provided !

I applied L'Hopital's rule, and it works very well (if I didn't miscalculate). For those who will find this thread in the future : L'hopital's rule helps calculating limits when the indeterminate form is [imath]0/0[/imath] or [imath]infinity/infinity[/imath]. It is also called "Bernoulli's rule", and it says that :
[math]lim(x->a) f(x)/g(x) = lim(x->a) f'(x)/g'(x)[/math]Where [imath]a[/imath] can be any integer or -infinity or +infinity.

This is true as long as f(x) and g(x) are drivable and g'(a) does not equal 0.

When we apply this to our problem :

[math]f(x) = x.ln(1-1/x) = (ln(1-1/x)/(1/x)[/math]
We now have, as mentioned by BigBeachBanana, an indeterminate form [imath]0/0[/imath] and l'Hopital's rule is applicable.

[math](ln(1-1/x))' = 1/(x^2-x)[/math][math](1/x)'= -1/x^2[/math]
So we have :
[math]lim(x-> +infinity) f(x) = lim(x-> +infinity) (1/(x^2-x))/(-1/x^2) = lim(x-> +infinity) (-1)/(1-1/x)[/math]
[math]lim(x-> +infinity) f(x) = -1[/math]
This rule is very general and this I why I chose that one to solve the problem, even though lookagain's solution is very elegant. Thank you pka for posting another way to solve this. I'm sure it will help someone else in the future.