What is the limit of sinx / xsinxx as x goes to infinity?
3 Answers
By Squeeze Theorem, this limit is 0.
Explanation:
Since
and since
By Squeeze Theorem
Explanation:
For any
Random advanced footnote
-
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 atx=0 to be1 ).
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
-sum_(n=1)^oo 1/(n^2 pi^2)
But
(sin x)/x = 1-x^2/6+x^4/120-...
So:
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.
Explanation:
What happens to our function when
No matter what the input,
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,
Hope this helps!