MHB What is the limit of \( x^x \) as \( x \) approaches 0?

[karush]

$\displaystyle \lim_{{x}\to{0}}x^x$

Well this has been posted before, and there seems to be several ways to do it. Assume also, if one is familiar with some of theories it can be solved just by observation. The answer is $1$ but I would of guessed "does not exist"

However since $0^0$ is indeterminate then use L’Hospital’s Rule:

So we could go from $y=x^x$ or $\ln\left({y}\right)=x\ln\left({x}\right)$

Not sure which way to go with this, and since it is a common SAT problem it would be nice to see how it solve with just a few steps. Thanks ahead...

 

[Dethrone]

Hi karush, (Wave)

The best way is to take the logarithm of both sides, as you have said:

$\ln\left({y}\right)=x\ln\left({x}\right)$

What do you get when you take the $\lim_{{x}\to{0}}$ on both sides?

I should add that we could rewrite the RH side so L'hopital's is apparent:
$$\lim_{{x}\to{0}}\frac{\ln\left({x}\right)}{x^{-1}}$$

 

[karush]

I should add that we could rewrite the RH side as that we can use L'hopitals:
$$\lim_{{x}\to{0}}\frac{\ln\left({x}\right)}{x^{-1}}$$

Ok if L'hopitals Rule is $\lim_{{x}\to{0}}\frac{f(x)}{g(x)}=\lim_{{x}\to{0}}\frac{f'(x)}{g'(x)}$

and $\d{}{x}\ln\left({x}\right)=x^{-1}$ and $\d{}{x}{x}^{-1}=-x^{-2}$

I don't see how we could get and answer of $1$ from this

 

[Dethrone]

That's right. Now put your answers together:

$$\lim_{{x}\to0{}}\ln\left({y}\right)=\lim_{{x}\to{0}}\frac{\ln\left({x}\right)}{x^{-1}}$$

$$\lim_{{x}\to0{}}\ln\left({y}\right)=\lim_{{x}\to{0}}\frac{(1/x)}{-(1/x^2)}$$

What do you get when you simplify? :D

 

[ineedhelpnow]

$x^x=e^{x*ln(x)}$

$\lim_{{x}\to{0}} e^{x*ln(x)}$

$\lim_{{x}\to{0}} x*ln(x) =0$ I am sure you know how to do this

$e^0=1$

 

[Dethrone]

$x^x=e^{x*ln(x)}$

$\lim_{{x}\to{0}} e^{x*ln(x)}$

$\lim_{{x}\to{0}} x*ln(x) =0$ I am sure you know how to do this

$e^0=1$

You spoiled my fun (Tmi)

$\lim_{{x}\to{0}} x*ln(x) =0$ => requires L'hopital's rule, which is what the OP is working on now.

Spoiler

P.S you went from ineedhelpnow to igivehelpnow. Name change suggested. ;)

 

[karush]

well just simplifying the $$\frac{1/x}{1/x^2}=-x$$

not sure about this next step but if $$\ln\left({y}\right)=-x$$

then $y=e^{-x}$ and when $x\to0$ then $y=1$

I saw the other posts but is this :)

 

[ineedhelpnow]

i don't want to spoil your fun. tell him if he did it right Rido.

- - - Updated - - -

>_< and i have gave help before

 

[Dethrone]

i don't want to spoil your fun. tell him if he did it right Rido.

Yes Ma'am (Nod)

(Turns towards Karush)

That is correct! (Muscle)

Here's how I would do it:
$$\lim_{{x}\to{0}}y=\displaystyle \lim_{{x}\to{0}}x^x$$
$$\lim_{{x}\to{0}}\ln\left({y}\right)=\lim_{{x}\to{0}}\frac{\ln\left({x}\right)}{x^{-1}}$$
$$\lim_{{x}\to{0}}\ln\left({y}\right)=\lim_{{x}\to{0}}\frac{(1/x)}{-(1/x^2)}$$
$$\lim_{{x}\to{0}}\ln\left({y}\right)=\lim_{{x}\to{0}}-x=0$$
$$\lim_{{x}\to{0}}y=e^0=1$$

You can also do it ineedhelpnow's way, in which we rewrite the limit:

$$\displaystyle \lim_{{x}\to{0}}x^x=\lim_{{x}\to{0}}e^{\ln\left({x^x}\right)}=\lim_{{x}\to{0}}e^{x\ln\left({x}\right)}=e^{\lim_{{x}\to{0}}x\ln\left({x}\right)}$$

Spoiler

ineedhelpnow, I'm just teasing (Sadface)

Spoiler

(Wait)(Devil)

 

[karush]

wow, that was help with some fireworks

Not sure tho who won the quick and dirty method..

 

[ineedhelpnow]

dont tell him but i did (Smirk)