Why does limx-->0+ 1/sin(x) equal x?

Question

Looking at the graph of sin(x), I thought sinx would approach 0 as x-->0. Same w/ tan(x). I don't understand why 1/sin(x) approaches x.

Answer 1

No., when x -> 0+, 1/sin(x) -> oo and when x -> 0-, 1/sin(x) -> -oo. This follows from the fact that sin(x) -> 0+ as x-> 0+ and sin(x) -> 0- when x -> 0-

It doesnt' make sense to say a limit approaches x or any other function. If the limit exists, it's either a real number or oo or -oo.

Answer 2

The answer is 0 not 1

Answer 3

As x->0, sin x doe sapproach 0, so 1/(sin x) goes to infinity. The answer to the limit has to either be a number or infinity (or be undefined). It cannot be a function (i.e. x).

Answer 4

it doesn't

sinx ~ x - x^3/6 + O(x^5)
1/sinx ~ (1/x)(1-x^2/6 +O(x^4))^(-1) ~ 1/x + x/6 +O(x^3) -->oo