L Rule Finding Limits

[Silvanoshei]

\(\displaystyle \lim \limits_{x\rightarrow 0^{+}} \sin \left(x\right)^{x} ==> lny=xlnsin(x)\)
\(\displaystyle \frac{(ln sin(x)) dx}{(x^-1) dx}==-> \frac{\frac{1}{sin(x)}}{-x^{-2}}\)
\(\displaystyle \lim \limits_{x\rightarrow 0^{+}} ==> \frac{\frac{1}{0}}{0} => \frac{\infty}{0} = 0?\)

 

[Silvanoshei]

Wait, I have to return it back to x don't I?

\(\displaystyle ln y=0 ==> e^{0} = 1?\)

 

[JeffM]

\(\displaystyle \lim \limits_{x\rightarrow 0^{+}} \sin \left(x\right)^{x} ==> lny=xlnsin(x)\)
\(\displaystyle \frac{(ln sin(x)) dx}{(x^-1) dx}==-> \frac{\frac{1}{sin(x)}}{-x^{-2}}\)
\(\displaystyle \lim \limits_{x\rightarrow 0^{+}} ==> \frac{\frac{1}{0}}{0} => \frac{\infty}{0} = 0?\)

I at least cannot make heads or tails of what you are doing above. It sort of appears that you know the answer, but are just guessing how to get there.

Let's try this substitution.

\(\displaystyle \theta = x^x \implies sin(x^x) = sin( \theta )\ and\ \theta = 1\ if\ x = 0.\)

\(\displaystyle \displaystyle \lim_{\theta \rightarrow 1}sin( \theta ) = what?\)

 

[Silvanoshei]

Taking Sin(x)^x and finding it's limit as it approaches 0+. So an exponent of 0 makes this infinity.

So, take it equal to y = sinx^x, and ln both sides. lny = ln sinx^x then you can use log property to shove the exponent out front.

Flip the x to 1/x on the bottom, so you have ln sinx / (1/x) or rather ln sinx / x^-1 ....

Use L Rule to d/dx... so my question is if I cleaned up ok with the ln y and come out with the correct value.

So.. \(\displaystyle y = e^{0} ?\)

 

[JeffM]

Taking Sin(x)^x and finding it's limit as it approaches 0+. So an exponent of 0 makes this infinity.

\(\displaystyle u^0 = 1,\ not\ \infty.\)

So, take it equal to y = sinx^x, and ln both sides. lny = ln sinx^x then you can use log property to shove the exponent out front.

Flip the x to 1/x on the bottom, so you have ln sinx / (1/x) or rather ln sinx / x^-1 ....

Use L Rule to d/dx... so my question is if I cleaned up ok with the ln y and come out with the correct value.

So.. \(\displaystyle y = e^{0} ?\)

I thought I explained this one. As I said in my response in the other thread, I have to leave so you will have to wait for someone else to try a different explanation. I don't get why you want to use logs.

 

[Silvanoshei]

I thought I explained this one. As I said in my response in the other thread, I have to leave so you will have to wait for someone else to try a different explanation. I don't get why you want to use logs.

L Rule = L'hopital's Rule. Sorry if the short version confused you. Evaluating the limit with this rule, u substitution doesn't fill those shoes to the correct answer. U-Sub works wonders though I agree. Anyone else know if my work is solid?

 

[Silvanoshei]

I thought this was true.... ?

\(\displaystyle sin(0)^{0} = \infty\)

 

[lookagain]

\(\displaystyle \lim \limits_{x\rightarrow 0^{+}} \sin \left(x\right)^{x} ==>\)

Silvanoshei, if what you are taking the limit of is \(\displaystyle \ sin(x^x), \ \) then type that, not what you did with the x as the exponent on the outside of the parenthesis.

Let's try this substitution. \(\displaystyle \theta = x^x \implies sin(x^x) = sin( \theta )\ \)\(\displaystyle and \ \theta = 1\ if\ x = 0. \ \ \ \ \ \)No, the limit as x --> 0+ of (x^x) = 1. \(\displaystyle \displaystyle \lim_{\theta \rightarrow 1}sin( \theta ) = what?\)

I thought this was true.... ?

\(\displaystyle sin(0)^{0} = \infty\)

Silvanoshei, don't type that. Type \(\displaystyle \ sin(0^0) \ \) if you mean that. Anyway, 0^0 is indeterminate.

Edit: Or, if you intend to take the limit of \(\displaystyle \ [sin(x)]^x, \ \) then put grouping symbols around the "sin(x)" part.