Evaluate limit [tln(t)] as t approaches 0

Question

Evaluate limit [tln(t)] as t approaches 0

Answer 1

Have you learned l'Hopital's rule yet?

 

l'Hopital's rule says that the limit of f(x) / g(x) is the same as the limit of f'(x) / g'(x), assuming the derivatives exist. and the limit of f'(x) / g'(x) exists.  But to use it you need to be evaluating the limit of a quotient.

 

Well, you can express this function as a quotient:  t ln t = ln t / (1/t) = ln t / t^-1

 

So in applying l'Hopital's rule to this, f(t) = ln t and g(t) = t^-1.

 

So f'(t) = 1/t = t^-1 and g'(t) = -1t^-2 = -t^-2.

 

So f'(t) / g'(t) = t^-1 / -t^-2 = -t.

 

The limit of -t as t approaches 0 is 0, so that's the limit of the original function.

 

 

Answer 2

As t → 0, ln(t) → - infinity. However, 0*anything = 0,

so the ANSWER IS ZERO

Answer 3

Since lnt is undefined when t ≤ 0, we only need to consider the limit as t approaches 0 from the right.

 

As t→0⁺, t →0 and lnt →-∞.  But, (0)(-∞) is an indeterminate form.  So, we need to put t lnt in a more manageable form.   

 

Rewrite tlnt as lnt/(1/t)

 

So, lim [lnt/(1/t)] as t → 0⁺ has the form -∞/∞, which is a situation where L'Hopital's Rule may be of help.    

 

Applying L'Hopital's Rule, the given limit is the same as                                 

       lim [(1/t)/(-1/t²)] as t →0⁺ = lim[-t] as t → 0⁺ = 0