Posted by Chelsea on Thursday, October 14, 2010 at 4:46am.
Are these correct?
lim x>0 (x)/(sqrt(x^2+4)  2)
I get 4/0= +/ infinity so lim x>0+ = + infinity? and lim x>0 = + infinity?
lim x>1 (x^2  5x + 6)/(x^2  3x + 2)
I get 2/0, so lim x> 1+ =  infinity? and lim x>1 = + infinity?
lim h>0 [(7)/(2+h^2) + (7/4)]/h
I used a computational website to get (7/4) as the answer, but I did not get this. My work ends with: (28)/(4(h+8) + (7(h+8)/(4(h+8)) and I end with 5.25 for an answer?
Last one!
lim x> neg. infinity (2x^2 + 3x  2)/(5x^3 + 4x x + 1)
Don't even know about this one?
Please help. I would like to understand these.

Calc Please Help  Reiny, Thursday, October 14, 2010 at 9:10am
The first two are correct.
Looking at the pattern of the third and considering the answer , I think you have a typo
The function appears to be f(x) = 7/x^2 and you are finding the derivative when x = 2
Instead of lim h>0 [(7)/(2+h^2) + (7/4)]/h , it should be
lim h>0 [(7)/(2+h)^2 + (7/4)]/h , notice the change in brackets.
then I get [28 + 7(2+h)^2]/(4(2+h)^2) / h
= [ 28 + 28 + 28h + 7h^2]/(4(2+h)^2) / h
= h(28+7h)/(4(2+h)^2) / h
= (28+7h)/(4(2+h)^2
now as h > 0 this becomes
28/16 = 7/4
for the last one, how about dividing each term in both the numerator and denominator by the highest power of x that you see, that is , by x^3
to get the expression as
(2/x + 3/x^2  2/x^3)/(5 + 4/x^2  1/x^2 + 1/x^3)
Now consider each term
As x becomes hugely negative, each of the terms would still be negative, but very very close to zero, so the top approaches zero, the bottom obvious approaches 5
so you have 0/5 which approaches 0 from the negative.