Question

Find the `lim_(x rarr oo) sinx/x` help me?

Answer 1

`lim_(x to oo) sin(x)/x=0`

Explanation:

What do we know about the function sin(x)?
It's a function, where `x in RR -> f(x) in [-1;1]`
So, you divide a bounded function such as sin(x) by x when x take a great value, the result is a very small value.
More generally, any bounded function divide by x give 0 in `+oo`
So : `lim_(x to oo) sin(x)/x=0`
\0/ here's our answer!

Answer 2

`lim x-> oo sin x/x=0`

Explanation:

As we have `-1<=sin x<=1` for all values of x and `|1/x||->0` as `x-> oo`
(where `|1/x|` means the absolute value of `1/x`),

it follows that `lim x-> oo |sin x/x|<=lim x-> oo 1/x=0`

Answer 3

`Lim_(x to oo) sin(x)/x=0`

Explanation:

`lim_(x to oo) sin(x)/x`
Here, we will use the squeeze theorem. We know that:
`-1<=sin(x)<=1`
`-1/x<=sin(x)/x<=1/x`
Because `lim_(x to oo)-1/x=lim_(x to oo) 1/x=0`, by theorem, `lim_(x to oo) sin(x)/x=0`
\0/ here's our answer!

Answer 4

`lim_(x rarr oo) sinx/x=0`

Explanation:

The bounds of the sine function are `[-1,1]`, meaning the numerator will just oscillate between those values.

The denominator will be the dominant term in this expression. As `x rarr oo`, `x` will just balloon up, becoming ridiculously large.

As the numerator stays around the same area, and the denominator will be much larger, we will be dividing by progressively larger numbers, yielding infinitesimally smaller numbers. Thus,

`lim_(x rarr oo) sinx/x=0`

Hope this helps!