Help pls!! 2 limit questions, feel free to answer both or either one!

[takelight]

Let \(\displaystyle y= (sin(x))^x\)

Then \(\displaystyle lny= xln(sin(x))\)

So \(\displaystyle \lim_{x\to 0} lny =\lim_{x\to 0} xln(sin(x))\)

Can you evaluate the right hand limit?

Um... that's just the same thing I have written differently. I showed my attempt to solve that limit in my comment above. I tried 3 l'hopitals and none worked.

 

[Deleted member 4993]

Please read:

http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx

Work the example problems - step-by-step.

 

[takelight]

L'Hopital's rule is not flawed at all. The rule says that each limit you get after applying L'Hopital's rule will be equal. L'Hopital's rule NEVER said anything about getting a limit that will be easy to compute. Do you understand the difference?

I meant my method of approaching the problem. sorry for the confusion.

 

[takelight]

Yes. I used it a couple more times and nothing. Why does it not work though, is the method flawed?

Here are my tries:

-(cosx)x²/sinx plug in the 0...…… 0/0

once more.... ((sinx)x² - 2xcosx)/cosx …… plugging in the 0 I get 0/1 which even gives 0 but I know the answer is 1. So where am I messing up?

WAIT.... I FORGOT THAT IT WAS e^ whatever I got. And so if I got 0 I just forgot that I had to plug It in e. So its e^0 which is 1 ok nice.

 

[Steven G]

Yes. I used it a couple more times and nothing. Why does it not work though, is the method flawed?

Here are my tries:

-(cosx)x²/sinx plug in the 0...…… 0/0

once more.... ((sinx)x² - 2xcosx)/cosx …… plugging in the 0 I get 0/1 which even gives 0 but I know the answer is 1. So where am I messing up?

You are confusing yourself! You evaluated a different limit, remember??
e^(x ln(sinx)), so lim x--> 0+ sinx^x is simply e^[whatever x(ln(sinx)) gives me when I check the lim] = e^[0] = 1!!

 

[Steven G]

Can someone please show me a limit that satisfies L'Hopital's requirement but does not yield an easier limit eventually? Thanks

 

[tkhunny]

When you plug the limit values into the functions that's exactly what you get.
For a, plug in 0
for b, 0^infinty

a) No. Never do that. Do you KNOW it's continuous at x = 0?
b) Very odd. Did you "plug in" [math]\infty[/math]?

 

[tkhunny]

Can someone please show me a limit that satisfies L'Hopital's requirement but does not yield an easier limit eventually? Thanks

How about [math]\lim_{x\rightarrow 0^{+}}x\cdot ln(x) = \lim_{x\rightarrow 0^{+}}\dfrac{ln(x)}{1/x}[/math]

 

[Steven G]

How about [math]\lim_{x\rightarrow 0^{+}}x\cdot ln(x) = \lim_{x\rightarrow 0^{+}}\dfrac{ln(x)}{1/x}[/math]

I am not sure about that.
[math]\lim_{x\rightarrow 0^{+}}x\cdot ln(x) = \lim_{x\rightarrow 0^{+}}\dfrac{ln(x)}{1/x} = \lim_{x\rightarrow 0^{+}}-x = 0[/math]

 

[tkhunny]

I am not sure about that.
[math]\lim_{x\rightarrow 0^{+}}x\cdot ln(x) = \lim_{x\rightarrow 0^{+}}\dfrac{ln(x)}{1/x} = \lim_{x\rightarrow 0^{+}}-x = 0[/math]

Aw! You simplified it after the first derivative application. Resist the temptation.