How do you find the limit of #(sinx)/(3x)# as x approaches #oo#?

Question

Isaiah Allen · Accepted Answer

0

you can see sinx as a wave picture whose value is always between 1 and -1 graph{sinx [-10, 10, -5, 5]} and then see 3x when x becomes larger and larger in denominator we dont need to care the value of sinx it will become 0

Christopher Allers · Answer

#lim_(xto +-oo) sin(x)/(3x)=0#

#color(blue)("Consider "sin(x) )#

This function can assume all values between and including -1 and +1. It will repeat this cycle every #2pi# radians as appropriate for the value of #x#. So #sin(x) in (-1 , +1)#
Where #(-1,+1)# is the range of all value between and including -1 to +1.

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider "3x)#

As #x# becomes increasing larger in the positive or negative direction, #1/(3x)# becomes increasingly smaller.

So for #x<0" "sin(x)/(3x)" tends to 0 but on the negative side of 0"#

So for #x>0" "sin(x)/(3x)" tends to 0 but on the positive side of 0"#

Thus #lim_(xto +-oo) sin(x)/(3x)=sin(x)/oo = 0#

As you move further and further away from the origin the amplitude lessens and approaches #y=0#

Christopher Allers · Answer

undefined To find the limit of (sinx)/(3x) as x approaches infinity, we can use the concept of limits. By applying the limit definition, we divide both the numerator and denominator by x. This gives us (sinx)/(3x) = (sin(x)/x)/(3).

Now, as x approaches infinity, the term sin(x)/x approaches 0, as the sine function oscillates between -1 and 1 while x grows without bound. Therefore, the limit of (sinx)/(3x) as x approaches infinity is 0/3, which simplifies to 0.