- #1
johnhuntsman
- 76
- 0
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.
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.