Understanding L'Hopital's Rule: Solving Limits with Indeterminate Forms

[DeadOriginal]

I recently learned about L'Hopital's Rule in calculus and as I was doing some practice problems I came across two that confused me a lot. I was hoping that someone could help me with them here.

The problem is lim_(t→0)⁡〖(sint)(lnt)〗.
I tried to make (sint)(lnt) a quotient by setting the problem up as: lim_(t→0)⁡〖(lnt)/(csct)〗
This made it so that both the top and the bottom of the function undefined when I plug in zero.

I thought maybe I would be able to differentiate both the top and the bottom and see if maybe that would lead me somewhere but it only made it worse by coming out as lim_(t→0)⁡〖x^(-1)/(-csctcott)〗
I always thought that because sin(0) is equal to 0 and since 0* anything is equal to 0 even if ln(0) is undefined then the limit of this equation is equal to 0. I don’t really trust that reasoning even though it sounds reasonable because it came out of my own head haha, and I’m not sure how I would show that as my work on a test so could you please give me some tips on how to proceed?

I also had a problem with lim_(x→∞)⁡〖x^(1/x) 〗
The way I did it was I took the function x^(1/x) and set it = y. I then took the ln of both sides so that I could make it lny = 1/xlnx
I then plugged that back into the limit to get:
lim_(x→∞)⁡〖lny= lim_(x→∞)⁡〖ln⁡(x^(1/x) )= lim_(x→∞)⁡〖1/x ln⁡〖x= 1/∞〗 〗 〗 〗 ln∞=0∞=0
I then used e to find the limit by:
lim_(x→∞)⁡〖x^(1/x)= lim_(x→∞)⁡〖e^(ln⁡(x^(1/x)))= e^lim_(x→∞)⁡〖ln⁡(x^(1/x))〗 = e^0=1〗 〗
If this isn’t the right way to do the problem, could you please explain to me how to do it?

Thanks,

Aaron Wong

WOW. I just looked over my post and it came out all funky. I'll try to clear things up if anything isn't understandable but if someone could point me in the right direction on how to post equations on the forum I'd be very grateful. Thanks.

 

[flyingpig]

This made it so that both the top and the bottom of the function undefined when I plug in zero.

Right there is the problem. L'Hopital's rule say that if the bottom function's derivative is 0, you can't apply the rule.

I suggest using ln^-1x

 

[DeadOriginal]

Right there is the problem. L'Hopital's rule say that if the bottom function's derivative is 0, you can't apply the rule.

I suggest using ln^-1x

Thanks for the quick reply.

So are you saying I should make the function sint/lnt instead of lnt/csct? Wouldn't that still come out with lnt undefined when I plug in 0?

 

[flyingpig]

Actually you know I misread your question. You could do it your way, but it wuold take you a long time...

Forget what i said about the bottom being 0 and stuff.

EDIT:I recommend you to have sin(x) at the top still

 

[HallsofIvy]

If, after taking the derivative of both numerator and denominator, you still get limits of 0 for both, use L'Hopital again. If the denominator now goes to 0 but the numerator does not, the limit does not exist (goes to infinity). If the numerator now goes to 0 but the denominator does not, the limit is 0, of course.

 

[DeadOriginal]

If, after taking the derivative of both numerator and denominator, you still get limits of 0 for both, use L'Hopital again. If the denominator now goes to 0 but the numerator does not, the limit does not exist (goes to infinity). If the numerator now goes to 0 but the denominator does not, the limit is 0, of course.

Thanks for the response.

So in this case, would undefined mean either inf or 0? Because when I plug in 0 to lnx it becomes undefined. I was always taught that no matter what undefined is wrong but I could have sworn that I read somewhere that ln(0) is equal to -inf so would it mean in this case that I did it wrong or that it is -inf.

If it is -inf then I could just make the equation lnx/(1/sinx). If 1/0 which is also undefined means inf then I would have inf/inf and I would be able to use L'Hopital's Rule. Would that be how it works? This is a little confusing... *scratches head*