Question

Explain how the limit of sinx as x approaches infinity does not exist

Answer 1

wait...is it because the range of sin only goes between -1 and positive 1?

Answer 2

are you allowed to use a graphing calculator

Answer 3

no

Answer 4

I guess I'm not really sure why it doesn't exist, because infinity goes on forever, so we don't have an exact value?

Answer 5

\(sin (x) = x -\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}-\cdots\)

\(\dfrac{sin(x) }{x}=1-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}-\cdots\)

as x goes to infinitive, can you tell me where sinx/x goes?

Answer 6

:O

Answer 7

oh i think I know why, its because the limit is a y point

Answer 8

and functions of sin go up and down forever (since its until infinity), so there is no clear y value

Answer 9

have you learned about limits approaching from the left and from the right?

Answer 10

yes

Answer 11

well i guess you can't really do that from the right