# limits involving x-root functions

• Nov 9th 2009, 07:33 PM
Landyach
limits involving x-root functions
Hey... This question is beyond me...

find lim as x -> 0

f(x) = 1 / ((x+1)^(1/x)) - 1/x

any help would be greatly appreciated.
• Nov 9th 2009, 08:01 PM
Bacterius
$\displaystyle f(x) = \frac{ 1}{(x+1)^{1/x}} - \frac{ 1}{x}$

You want :

$\displaystyle \lim_{x\to 0} f(x)$

Therefore, you want :

$\displaystyle \lim_{x\to 0} f(x) = \frac{ 1}{(x+1)^{1/x}} - \frac{ 1}{x}$

Let's not forget that $\displaystyle \lim_{x\to 0}\frac{ 1}{x} = \pm \infty$ (depending on which side you come from)

Consider if you are approaching $\displaystyle 0$ on the positive side :

$\displaystyle \lim_{x\to 0} f(x) = \frac{ 1}{1^{(+\infty)}} - (+\infty)$

Ugly, eh ?

$\displaystyle \lim_{x\to 0} f(x) = 1 - \infty$

Which by absorption (I'm not sure about the last part) :

$\displaystyle \lim_{x\to 0} f(x) = - \infty$

The other possibility (approaching $\displaystyle 0$ on the negative side) seems quite harder (since you get some weird results) but since I'm not really sure about what I've just done, I won't go on.
• Nov 9th 2009, 08:05 PM
Prove It
Quote:

Originally Posted by Bacterius
Let's not forget that $\displaystyle \lim_{x\to 0}\frac{ 1}{x} = \pm \infty$ (depending on which side you come from)

This would mean $\displaystyle \lim_{x \to 0}\frac{1}{x}$ does not exist.
• Nov 9th 2009, 08:07 PM
Bacterius
What do you mean ? My maths teacher told us the inverse of x could have two different limits in 0 depending if you approach 0 from the negative side (the limit is -infinity) or the positive side (the limit is +infinity).
Was he wrong or was I just doing something else when he said this ?

EDIT : now that I think about it I remember he said something about 0, maybe he said $\displaystyle \lim_{x \to 0}\frac{1}{x} = 0$ ?
• Nov 9th 2009, 08:29 PM
Prove It
Quote:

Originally Posted by Bacterius
What do you mean ? My maths teacher told us the inverse of x could have two different limits in 0 depending if you approach 0 from the negative side (the limit is -infinity) or the positive side (the limit is +infinity).
Was he wrong or was I just doing something else when he said this ?

EDIT : now that I think about it I remember he said something about 0, maybe he said $\displaystyle \lim_{x \to 0}\frac{1}{x} = 0$ ?

1. Please note my edit, I wrote the wrong value for x to go to.

2. He is not wrong, but if the function approaches the same value from the right as it does from the left, it does not exist.

However, the left hand limit is $\displaystyle -\infty$ and the right hang limit is $\displaystyle \infty$.
• Nov 9th 2009, 08:37 PM
Landyach
Quote:

Originally Posted by Bacterius
$\displaystyle f(x) = \frac{ 1}{(x+1)^{1/x}} - \frac{ 1}{x}$

You want :

$\displaystyle \lim_{x\to 0} f(x)$

Therefore, you want :

$\displaystyle \lim_{x\to 0} f(x) = \frac{ 1}{(x+1)^{1/x}} - \frac{ 1}{x}$

Let's not forget that $\displaystyle \lim_{x\to 0}\frac{ 1}{x} = \pm \infty$ (depending on which side you come from)

Consider if you are approaching $\displaystyle 0$ on the positive side :

$\displaystyle \lim_{x\to 0} f(x) = \frac{ 1}{1^{(+\infty)}} - (+\infty)$

Ugly, eh ?

$\displaystyle \lim_{x\to 0} f(x) = 1 - \infty$

Which by absorption (I'm not sure about the last part) :

$\displaystyle \lim_{x\to 0} f(x) = - \infty$

The other possibility (approaching $\displaystyle 0$ on the negative side) seems quite harder (since you get some weird results) but since I'm not really sure about what I've just done, I won't go on.

Wow... looking at it that way makes it simple... thanks man.

Also for the negative side its still easy...

1/1^anything = 1 therefore answer is 1 - 1/x = 1 +infinity = infinity

(or so I think)
• Nov 9th 2009, 08:48 PM
Bacterius
Yes you are right, it is basically the same thing on the other side, my mistake.