Question

What is `lim_(xtooo)(sin2x)/x`?

Answer 1

`0`

Explanation:

Think of this as:
`lim_(xtooo)sin2x*lim_(xtooo)1/x`

As `x` tends to `oo`, `1/x` gets smaller and smaller, until it is effectively `0`

However, `sin(2x)` can be be anywhere in the range `[-1,1]` at infinity.

So, we have: `lim_(xtooo)sin2x*lim_(xtooo)1/x=[0(-1),0(1)]=0`