# another limit

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• Nov 11th 2010, 01:52 PM
Vamz
another limit
$\displaystyle \lim_{x \to 0^{+}} x^{1/3} \ln x =$

How do I go about solving this kind of limit?
Do I handle each part individually?

The only thing I can think about, is taking the derivative of each part.

$
\displaystyle \frac{1}{3x^{\frac{2}{3}}} * \frac{1}{x}
$

To me, this still looks like it is undefined..
Can someone please explain how to do these types? Perhaps a link to a tutorial too. I'm not sure what to search online for these types either. Do they have a special name?

Thanks!
• Nov 11th 2010, 02:02 PM
dwsmith
$\displaystyle\lim f*g=\lim f *\lim g$

$\displaystyle \lim_{x\to 0}x^{\frac{1}{3}}=0$

$\displaystyle 0*\lim_{x\to 0}ln(x)=0$

A useful tool for some limits is:

$\displaystyle \lim_{x\to\infty}f(x)=L \ \mbox{iff.} \lim_{x\to 0}\frac{1}{f(x)}=L$

Using this on your limit, we get:

$\displaystyle \lim_{x\to\infty}\frac{1}{x^{\frac{1}{3}}*ln(x)}=\ lim_{x\to\infty}\frac{1}{\infty}=0$
• Nov 11th 2010, 02:20 PM
Vamz
Thank you!

So, since the limit of f(x) =0, we can ignore whatever the limit of g(x) is, since we are multiplying. But for this scenario, would it be safe to say that the limit of g(x) would have been undefined? In that case 0 * undefined = 0?

Thanks for that tool!
However, did you mean to say:
$
\displaystyle \lim_{x\to 0}f(x)=L \ \mbox{iff.} \lim_{x\to \infty}\frac{1}{f(x)}=L
$

did you have the 0 & infinity symbols backwards?
• Nov 11th 2010, 02:25 PM
dwsmith
You can't just ignore anything and no the 0 and infinity aren't backwards.

For example, lets consider the function $\displaystyle f(x)=\frac{1}{x}$ then the $\displaystyle\lim_{x\to\infty}\frac{1}{x}=0$

Ok so lets use the definition now. $\displaystyle\frac{1}{f(x)}=\frac{1}{\frac{1}{x}}= x$

$\displaystyle\lim_{x\to 0}{x}=0$
• Nov 11th 2010, 02:57 PM
Vamz
Ok IC. So for
$
\displaystyle \lim_{x \to 1^{+}} ( \displaystyle \frac {1}{\ln x} - \displaystyle \frac {1}{x - 1} ) =
$

It would be the same as: Lim f(x) - Lim f(g)
so for:
$
\lim_{x \to 1^{+}}\frac{1}{lnx} = \infty
$

x becomes really close to one, making ln(x) extremely small. Then 1/rly small = rly big and approaches infinity?

Same with the other term:
$$\displaystyle \lim_{x \to 1^{+}} \frac{1}{x-1} = \infty$$

because x-1 will approach a super small number, and the reciprocal of that will be a rly big number, approaching infinity.

So, then we have infinity-infinity = 0?

But this appears to be wrong. What is wrong with my reasoning?
• Nov 11th 2010, 03:11 PM
dwsmith
You can't subtract infinities. Some infinities are bigger than others. You need to use L'Hopital's Rule here.
• Nov 11th 2010, 03:31 PM
skeeter
Quote:

Originally Posted by Vamz
Ok IC. So for
$
\displaystyle \lim_{x \to 1^{+}} ( \displaystyle \frac {1}{\ln x} - \displaystyle \frac {1}{x - 1} ) =
$

It would be the same as: Lim f(x) - Lim f(g)
so for:
$
\lim_{x \to 1^{+}}\frac{1}{lnx} = \infty
$

x becomes really close to one, making ln(x) extremely small. Then 1/rly small = rly big and approaches infinity?

Same with the other term:
$$\displaystyle \lim_{x \to 1^{+}} \frac{1}{x-1} = \infty$$

because x-1 will approach a super small number, and the reciprocal of that will be a rly big number, approaching infinity.

So, then we have infinity-infinity = 0?

But this appears to be wrong. What is wrong with my reasoning?

$\infty - \infty$ is an indeterminate form.

$\displaystyle \frac{1}{\ln{x}} - \frac{1}{x-1}$

$\displaystyle \frac{(x-1) - \ln{x}}{\ln{x}(x-1)}$

use L'Hopital ...

$\displaystyle \frac{1 - \frac{1}{x}}{\ln{x} + \frac{x-1}{x}}$

multiply numerator and denominator by $x$ ...

$\displaystyle \frac{x - 1}{x\ln{x} + (x-1)}$

use L'Hopital again ...

$\displaystyle \frac{1}{2 + \ln{x}}$

limit as $x \to 1^+$ is $\frac{1}{2}$
• Nov 11th 2010, 03:56 PM
Vamz
There we go. I would have never thought of multiplying x on both the numerator and denominator :s

thanks a ton!