Struggling with Calculus Limits?

[J-Girl]

Limits- need help!

Hi all, I am currently doing calculus in maths and finding it difficult, keep going through the material but its not making sense and not providing much explanation on solved problems. Can anybody help me with a few questions, i have attempted them but would pleasepleaseplease like somebody to check them and tell me if I am wrong, and a solution??

Evaluate the following limits, if they exist:
a) lim
x--> -2 (x^3 + x^2 -x+2)
=(-2)^3 + (-2)^2 -(-2) + 2
=-8 + 4 + 2 + 2
=0, so the limit of f(x)=0

b)lim
x--> -1 (x+3)/|x-2|
pos= (-1+3)/(-1-2)
= 2/-3
neg= (x+3)/-(x-2)
= (x+3)/(-x--2)
=(x+3)/(-x+2)
=(-1+3)/(--1+2)
=2/(1+2)
=2/3
so, since there is no unique value (-2/3 and 2/3) the limit does not exist

c)lim
x-->2- (x+3)/|x-2|
neg= (x+3)/-(x-2)
= (x+3)/(-x--2)
=(x+3)/(-x+2)
=(2+3)/(-2+2)
=5/0
= no limit

d) lim
x-->0 ((x+4)^2-16)/4x
=x^2+4x+4x+16-16
=x^2 +8x
=(x^2 + 8x)/4x
=(x+8)/4
=(0+8)/4
=8/4
=2

e) lim
x-->-infinity (x^3 + x^2 + x -2)/(2-5x)^3
divide numerator and deniminator by x^3
=(x^3+x^2+x-2)/(8-60x+150x^2-125x^3)
dividing numberator & denominator by x^3
i got lim 1/lim -125, so limit is -1/125

id really appreciate if somebody could please just reassure me:) thanx!

 

[Dick]

The first problem I see is in b). If x is a number close to -1, then x-2 is negative. There is no need to consider a positive case.