L'Hospital's problem, help please?

Question

this is a calculus II problem thats I really don't get.

lim (lnx)^(sinx)
x ->1

The limit tends to 1 really confuses me. Any kind soul willing to help?

I understand limits and continuity. But this problem came from my BC II book, in the L'hospital's section. I tried using L'H by taking the limits of both sides and ending up with:

ln y = lim __ln(lnx)__
x->1 1/sinx

But this is as fas as I can get, as I realize putting sinx as reciprocal wouldn't help. Is there any other way to use the rule and end up with 0 as my final answer?

Answer 1

This is where you really need to understand the difference between a point and a limit. The 'point' at coordinates [1, f(x)] does not exist since f(x) fails to exist at x=1. But, if it did, it would be at [1, 1].

Imagine yourself walking down a path in a garden. The path is set out with little, round 'paver' stones that are just touching. Suddenly, you come to a spot where one of the stones is missing. But you still know exactly where it would be---- if it were there.

That's the example I used to give my Calculus students. I know that it's a kinda poor one (since between any 2 points there are an uncountable infinity of 'other' points both rational and irrational), but it seemed to make them 'feel good' about the idea of a limit.

HTH

Doug

Answer 2

l'Hopital applies when you have a quotient, where both numerator and denominator go to zero at the limit point. You don't have that here; what you have is a mess. But you can do the problem directly; ln 1 is zero and sin 1 is 0.8414, and zero to any finite power is zero. So I make the limit to be zero.

Answer 3

If this ^ was ment to be an exponent, you cant apply L'Hopital here. You need a quotient for that.

Ana