Question

Solve:

Lim
x--> 0 sinx/x^2

Answer 1

first apply L hopital then normal limit rules

Answer 2

How would you solve this without L'hopitals?

Answer 3

at x--->0 it is undefined (asymptote at 0) but if you come from the left or right side it is
at x-->0- it is -infinity
at x-->0+ it is +infinity

Answer 4

How did you get this result? Can someone show it step by step?

Answer 5

im x -> 0 only exists if:

lim x -> 0+ = lim x -> 0-

[this implies that f(x) is continuous at 0]

to manually find THE LIMITS OF 0+ and 0- you can plug

0+ = 0.0000001
0- = -0.0000001

if you get the same result for both, limit exists and is the number you get.

if you don't. limit DNE

Answer 6

\(\dfrac{\sin(x)}{x^{2}} = \dfrac{\sin(x)}{x}\cdot \dfrac{1}{x}\)

It's a lot easier to look at it that way.

Answer 7

Should I be using squeeze theorem?

Answer 8

Yes, for the sin(x) / x you can use the squeeze theorem to prove the limit is 1 as x->0.

1/x -> -infinity x-->0-
1/x -> +infinity x-->0+