# Math Help - L'Hopital's Rule (a different problem)

1. ## L'Hopital's Rule (a different problem)

ex: ln lim x-> 0+ x^(sin x), y=x^(sin x)

using L'Hopital's Rule, the book says the answer is 1 but i don't know how to get to this answer

Thanks so much!

2. Looking at the individual functions, we see that both approach 0 as X approaches zero from the left. This would be an indeterminate power, of the form $0^0$ and thus we have to use L'hopital's rule.

To solve this we first take the natural logarithim of our function:

$\ln(y)=\ln(x)^{\sin(x)}$

Which then gives us:

$\ln(y)=\sin(x)\ln(x)$

This then becomes an indeterminate product of the form $0*-\infty$

Do you see what to do at this point?

3. it says latex error! syntax error

4. Sorry. I suck at this LaTex stuff. Finally figured it out. By accident.

5. i'm really sorry but how does lny = ln x ^sin x

6. No need to be sorry. Our properties of logarithims says:

$\ln(b)^a$ is equal to $a\ln(b)$. Because $\ln(b)^a$ is really $\log_{e}(b)^a$; and hopefully we know a logarithmic property that allows us to manipulate that.

7. hmm okay so first let me ask a question:

does ln lim x sinx = lim ln x sinx ?

and i did see how you got ln y= sin x ln x
from there i can say lim sinx ln x ?
x->0+

8. Originally Posted by ricey
hmm okay so first let me ask a question:

does ln lim x sinx = lim ln x sinx ?
No. Because this implies that $x=\ln{x}$. Obviously not true.

and i did see how you got ln y= sin x ln x
from there i can say lim sinx ln x ?
x->0+[/quote]

I'm not sure that I understand your question...

9. Hmmm, I believe he is asking if he can write

$\lim_{x \rightarrow 0^{+}} \sin(x) \ln(x)$ is equal to $\sin(x) \lim_{x \rightarrow 0^{+}}\ln(x)$, which is of course no. I'm thinking OP is applying limit rearrangment of the type:

$\lim_{x \rightarrow a} \sqrt[n]{f(x)}$ is the same as $\sqrt[n]{\lim_{x \rightarrow a}f(x)}$

The first way of writing it is how we want to set it up.

His second question is can he write it like this:
$\lim_{x \rightarrow 0^{+}} \sin(x) \ln(x)$

Which is the proper way to write it, but not using the logic he used. I apologize OP for not writing out the limit notation, as I just assumed you realized where they went.