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

3 Answers
Sep 28, 2015

By Squeeze Theorem, this limit is 0.

Explanation:

Since AAepsilon>0, => 0<=|sinx/x|<=|1/x| AAx>=1/epsilonε>0,0sinxx1xx1ε

and since lim0=0 and lim1/x=0

By Squeeze Theorem =>limsinx/x=0

Sep 28, 2015

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.

Aug 2, 2018

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!