Limits
limx->-3 (6+4x)/(x2+1)
-3/5
limt->-4sin(Pi*t)/(t2-16)
-Pi/8
limn->infinity 4n/(12n/4n)
diverges to infinity
limt->0 [(6+t)2-36]/h
12
limt→1
(5t^4−4t^2−1)/(10−t−9t^3)
-3/7
limn->infinity -7n/7n
diverges
limx->-5(x2-25)/(x2+2x-15)
5/4
Limx→infinity x3*e-x
0
limn->infinity 12n/18n
converges to 0
limx->-3 (sqrt(2x+22)-4)/(x+3)
1/4
limx->- infinity xex
0
limn->infinity 25n/25n
Prove that 1+3+5+...+(2n-1)=n2
See Micah
limx->0 x/(3-sqrt(x+9))
-6
limx->infinity x1/x
1
limn->infinity 3n+15/n
3
Prove 2+4+6+...+2n=n2+n
See Micah