## clara1223 one year ago find the limit as h approaches 0 of h(3+(2/h^2))

1. clara1223

$\lim_{h \rightarrow 0} h(3+\frac{ 2 }{ h ^{2} })$

2. amistre64

we can try some test points, say h=.0000001, and h=-.0000001 but what does your thoughts tell you initially?

3. clara1223

Well I solved and got that the limit is 2, but that was wrong.

4. amistre64

your solution was flawed then :) we have 2 terms: ab if h < 0, a=-, b=+ ... ab = - if h > 0, a=+,b=+ ... ab=+

5. clara1223

I don't understand.

6. amistre64

let a=h let b=3+2/h^2 then multiply them together and yo have your setup. when h is less than zero, what are our signs for a and b? what about their product?

7. clara1223

if h is less than 0 then a would be negative and b would be positive, and their products would both be negative

8. amistre64

ok, so we have a left side limit that is negative ... now assume h is greater than zero ... what are the signs of a and b? and of their product?

9. clara1223

if h is greater than 0 then a and b would both be positive and so would be their products.

10. clara1223

oh, so since from the left it's negative and from the right it's positive, it doesnt exist?

11. amistre64

so we have a right side limit that is positive can a positive limit ever be equal to a negative limit?

12. amistre64

correct, its actually -inf and +inf so yeah

13. clara1223

yeah, I graphed it and it helped me to understand.

14. amistre64

graphs do help if applicable :)