What is the limit of #sinx / x# 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 at#x=0# to be#1# ).
Hence by the Weierstrass factorisation theorem:
#(sin x)/x = (1x/pi)(1+x/pi)(1x/(2pi))(1+x/(2pi))(1x/(3pi))(1+x/(3pi))...#
#color(white)((sin x)/x) = (1x^2/pi^2)(1x^2/(4pi^2))(1x^2/(9pi^2))...#
Hence the coefficient of
#sum_(n=1)^oo 1/(n^2 pi^2)#
But
#(sin x)/x = 1x^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!