Find limit of 8t*ln(t) as t approaches 0 from the right.?

Question

I know we would use L'hopital's rule but i still can not figure out the answer yet. little help pls!

Answer 1

The answer is 0

Answer 2

L'Hopital's Rule applies to fractions, so let's make your function a fraction:

ln(t) / (1/8t)

The numerator will approach negative infinity and the denominator will approach positive infinity, so the limit will be negative or zero (approach zero from the left).

Now take the first derivative of the numerator and denominator.

The first derivative of the numerator is 1 / t.

The derivative of the fraction in the denominator is

- 8 / 64t^2 = -1 / 8t^2

If L'Hopital's Rule is going to give us anything good, we have

(1/t) / (-1/8t^2) as t approaches 0 from the right.

Cancelling, we have

1 / -1/8t, or -8t. The limit of the original function is thus zero, approached from the left.

It is plausible, since ln(t) approaches negative infinity and 8t approaches 0, so anything is possible.

James, below, ingeniously leaves out the 8, which is an unnecessary red herring, from his analysis. I'm glad a genius such as him agrees with my answer, and by the same method, too.

Answer 3

as t->0, 8t ->0, but ln(t) -> -infinity. So, in order to apply l'Hospital's Rule, you need to rewrite this as a quotient of functions that both go to 0 or both become infinite as t->0.

This can be accomplished by rewriting it as

ln(t) / (1/t),

both of which become infinite as t->0.

Now apply l'Hospital's Rule:

1/t / (-1/t^2)

and simplify to get -t, which goes to 0 as t->0.

Answer 4

I verified the answer with Maple(very handy!!) and it is indeed 0. You should buy yourself a copy or if you're in college I'm sure they have it at computing site or something and the method the second guy used is also 100% correct.

Answer 5

56

Answer 6

I too have the same question

Answer 7

Maybe, yeah