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

2 Answers
冠廷 李.
Jun 16, 2016

0

Explanation:

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

Tony B
Jun 16, 2016

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

Explanation:

#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#

Tony B

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

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