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!
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 $\displaystyle 0^0$ and thus we have to use L'hopital's rule.
To solve this we first take the natural logarithim of our function:
$\displaystyle \ln(y)=\ln(x)^{\sin(x)}$
Which then gives us:
$\displaystyle \ln(y)=\sin(x)\ln(x)$
This then becomes an indeterminate product of the form $\displaystyle 0*-\infty$
Do you see what to do at this point?
No need to be sorry. Our properties of logarithims says:
$\displaystyle \ln(b)^a$ is equal to $\displaystyle a\ln(b)$. Because $\displaystyle \ln(b)^a$ is really $\displaystyle \log_{e}(b)^a$; and hopefully we know a logarithmic property that allows us to manipulate that.
Hmmm, I believe he is asking if he can write
$\displaystyle \lim_{x \rightarrow 0^{+}} \sin(x) \ln(x)$ is equal to $\displaystyle \sin(x) \lim_{x \rightarrow 0^{+}}\ln(x)$, which is of course no. I'm thinking OP is applying limit rearrangment of the type:
$\displaystyle \lim_{x \rightarrow a} \sqrt[n]{f(x)}$ is the same as $\displaystyle \sqrt[n]{\lim_{x \rightarrow a}f(x)}$
The first way of writing it is how we want to set it up.
His second question is can he write it like this:
$\displaystyle \lim_{x \rightarrow 0^{+}} \sin(x) \ln(x)$
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.