Question

What is the limit of `sinx / x` as x goes to infinity?

Answer 1

By Squeeze Theorem, this limit is 0.

Explanation:

Since `AAepsilon>0, => 0<=|sinx/x|<=|1/x| AAx>=1/epsilon`

and since `lim0=0 and lim1/x=0`

By Squeeze Theorem `=>limsinx/x=0`

Answer 2

`lim_(x->oo) (sin x)/x = 0`

Explanation:

`0 <= abs((sin x) / x) <= abs(1/x) -> 0` as `x -> oo` (also as `x->-oo`)

For any `epsilon > 0`, we find `abs((sin x) / x) < epsilon` for all `x > 1/epsilon`.

Random advanced footnote

`(sin x)/x` has some interesting properties and uses:

• `lim_(x->0) (sin x)/x = 1`

• `(sin x)/x = 0 \ <=> \ x = kpi " for " k in ZZ " with " k!=0`

• `(sin x)/x` is an entire function. That is it is holomorphic at all finite points in the complex plane (taking its value at `x=0` to be `1`).

Hence by the Weierstrass factorisation theorem:

`(sin x)/x = (1-x/pi)(1+x/pi)(1-x/(2pi))(1+x/(2pi))(1-x/(3pi))(1+x/(3pi))...`

`color(white)((sin x)/x) = (1-x^2/pi^2)(1-x^2/(4pi^2))(1-x^2/(9pi^2))...`

Hence the coefficient of `x^2` in the Maclaurin expansion of `(sin x)/x` is:

`-sum_(n=1)^oo 1/(n^2 pi^2)`

But `sin x = x/(1!)-x^3/(3!)+x^5/(5!)-...`, so:

`(sin x)/x = 1-x^2/6+x^4/120-...`

So: `1/6 = sum_(n=1)^oo 1/(n^2 pi^2)` and hence:

`sum_(n=1)^oo 1/n^2 = pi^2/6`

Finding this sum was the Basel Problem, set by Pietro Mengoli in 1644 and solved by Leonard Euler in 1734.

Answer 3

`lim_(xtooo)sinx/x=0`

Explanation:

What happens to our function when `x` balloons up?

No matter what the input, `sinx` just oscillates between `0` and `1`.

As the denominator gets larger and larger, we will be dividing by a larger number, which yields a smaller number.

Since the numerator stays relatively the same, and the denominator blows up, `sinx/x` will become infinitesimally small and approach zero.

Hope this helps!