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

Question

Paisley Johnson · Accepted Answer

By Squeeze Theorem, this limit is 0.

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#

Christopher Allers · Answer

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

#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.

Charles Arocha · Answer

#lim_(xtooo)sinx/x=0#

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!

Elijah Abram · Answer

undefined The limit of sinx / x as x goes to infinity is 0.