1. ## limits question

lim x-->0

$\displaystyle \frac{{x}^{x} -1}{x lnx}$

also it says assume as x--> 0 xlnx --> 0

my answer is its 0/0 type since anything to the power of 0 is 1 including $\displaystyle {x}^{x}$ ?

using L H Rule you get

$\displaystyle \frac{{x}^{x} (1+lnx)}{(1+lnx)}$

then again the anything to the power of zero thing gives you that $\displaystyle {x}^{x}$ --> 1

2. ## Re: limits question

There is no such "thing". 0 to the 0 power is NOT 1. What L'Hopital's rule says is that the limit is the same as the limit of $\displaystyle x^x$. To find that limit, let $\displaystyle y= x^x$ so that $\displaystyle ln(y)= x ln(x)$. If you are allowed to "assume" the limit of that is 0, then the limit of ln(y) is 0 so the limit of $\displaystyle x^x$ is, indeed, 1.

3. ## Re: limits question

Originally Posted by Applestrudle
lim x-->0

$\displaystyle \frac{{x}^{x} -1}{x lnx}$

also it says assume as x--> 0 xlnx --> 0

my answer is its 0/0 type since anything to the power of 0 is 1 including $\displaystyle {x}^{x}$ ?

using L H Rule you get

$\displaystyle \frac{{x}^{x} (1+lnx)}{(1+lnx)}$

then again the anything to the power of zero thing gives you that $\displaystyle {x}^{x}$ --> 1
what you need to properly finish this is the fact that

$\large x^x = e^{x \ln(x)}$

you were told that $\displaystyle{\lim_{x \to 0}}~ x \ln(x) = 0$ so $\displaystyle{\lim_{x \to 0}}~e^{x \ln(x)} =e^0 = 1$

4. ## Re: limits question

It's not that difficult to show $\displaystyle \lim_{x\to 0} x\ln{x} = 0$. If you write $\displaystyle x\ln{x} = \frac{\ln{x}}{\frac{1}{x}}$, then using L'Hopital's rule gives $\displaystyle \frac{\frac{1}{x}}{-\frac{1}{x^2}}$, which is -x.

- Hollywood

5. ## Re: limits question

Originally Posted by hollywood
It's not that difficult to show $\displaystyle \lim_{x\to 0} x\ln{x} = 0$. If you write $\displaystyle x\ln{x} = \frac{\ln{x}}{\frac{1}{x}}$, then using L'Hopital's rule gives $\displaystyle \frac{\frac{1}{x}}{-\frac{1}{x^2}}$, which is -x.

- Hollywood
Actually the limit \displaystyle \begin{align*} \lim_{x \to 0} x\ln{(x)} \end{align*} DOES NOT EXIST as \displaystyle \begin{align*} x\ln{(x)} \end{align*} is only defined where \displaystyle \begin{align*} x > 0 \end{align*}. However, \displaystyle \begin{align*} \lim_{x \to 0^+} x\ln{(x)} = 0 \end{align*}.

Good point.

Thanks,
Hollywood