Solving for t->0+: L'Hospital's Rule

[johnhuntsman]

lim__[[t^3 ln(t)] / 3] + (t^3)/9
t->0+

^ in case that's difficult to read here's this.

Now what I don't understand is how this comes out to be 0. I would think that ln(t) as t approaches 0 would be -∞ (since ln(t) keeps getting bigger and bigger in the - direction as t approaches o). You would multiply that by 0 (since t^3 as t approaches 0 gets closer and closer to 0). 0 * -∞ should be indeterminate. 0 * -∞ over 3 (i.e., the term: [t^3 ln(t)] / 3]) should also be indeterminate. I understand how (t^3)/9 is 0. Is this problem meant to be solved with L'Hospital's Rule?

P.S. I've used L'Hospital's rule on it already and it does come out to zero. I just want to check with the people here to see if that is whatI'm supposed to do.

 

[STEMucator]

lim__[[t^3 ln(t)] / 3] + (t^3)/9
t->0+

^ in case that's difficult to read here's this.

Now what I don't understand is how this comes out to be 0. I would think that ln(t) as t approaches 0 would be -∞ (since ln(t) keeps getting bigger and bigger in the - direction as t approaches o). You would multiply that by 0 (since t^3 as t approaches 0 gets closer and closer to 0). 0 * -∞ should be indeterminate. 0 * -∞ over 3 (i.e., the term: [t^3 ln(t)] / 3]) should also be indeterminate. I understand how (t^3)/9 is 0. Is this problem meant to be solved with L'Hospital's Rule?

P.S. I've used L'Hospital's rule on it already and it does come out to zero. I just want to check with the people here to see if that is whatI'm supposed to do.

Yes, Hospital's rule is appropriate for this question. Something else you could've done was use the squeeze theorem.

 

[johnhuntsman]

Alright thanks. I appreciate it.

 

[Ray Vickson]

lim__[[t^3 ln(t)] / 3] + (t^3)/9
t->0+

^ in case that's difficult to read here's this.

Now what I don't understand is how this comes out to be 0. I would think that ln(t) as t approaches 0 would be -∞ (since ln(t) keeps getting bigger and bigger in the - direction as t approaches o). You would multiply that by 0 (since t^3 as t approaches 0 gets closer and closer to 0). 0 * -∞ should be indeterminate. 0 * -∞ over 3 (i.e., the term: [t^3 ln(t)] / 3]) should also be indeterminate. I understand how (t^3)/9 is 0. Is this problem meant to be solved with L'Hospital's Rule?

P.S. I've used L'Hospital's rule on it already and it does come out to zero. I just want to check with the people here to see if that is whatI'm supposed to do.

For any fixed power p > 0 we have
\lim_{x \rightarrow 0+} x^p \ln(x) = 0, so things like
x^{1/10} \ln(x), \; \sqrt{x} \ln(x),\; x \ln(x), \ldots
all → 0 as x → 0+. It might help to write
x^p \ln(x) = \ln \left(x^{x^p}\right) \text{ for } x > 0.

RGV