Question

How do I find the limit as `x` approaches negative infinity of `sinx`?

Answer 1

Limit does not exist

Explanation:

`Lim x-> -oo` sin x = `Lim x->oo` sin(-x)= `Lim x->oo` -sinx

As `x->oo`, sin x will oscillate between -1 and +1 and will not converge to any particular value. Hence this limit would not exist.