Find limit if exists

Question

1. Lim x goes to 0 x^2cos(1/x^3)

2.lim x goes to 1 to the left
4x/x^2+2x-3

3. Lim x goes to π to the right
cscx

Answer 1

#1 cos(1/x^3) oscillates between 1 and -1, but the x^2 out front forces the limit to zero

http://www.wolframalpha.com/input/?i=x^2cos%281%2Fx^3%29+

#2 for x < 1, x^2+2x-3 < 0
So, you have 4*1/(small negative) -> -∞

#3 same thing here. sin(x) > 0 for x<π, so csc(x) -> +∞

You can type in your other functions at wolframalpha to verify the limits.