Question

X is approaching infinity when the limit is equal to xsin(pi/2)

Answer 1

you mean evaluate the limit of x sin(pi/2) as x approaches infinity

just saying because I don't believe you mean the limit is x sin(pi/2)

Answer 2

\[\lim_{x \rightarrow \infty} x \sin(\frac{\pi}{2})\]
is this really what you want to evaluate?

Answer 3

because this is constant times a large large number

Answer 4

Yes it's what I want, it's my homework

Answer 5

yeah that should be the format ofthe equation

Answer 6

you do know sin(pi/2) is 1
so you have
\[\lim_{x \rightarrow \infty} x\]

Answer 7

you already know the answer
if x approaches infinity, then x approaches ?
:)

Answer 8

i dont think i should post the answer, right?

Answer 9

Omg it's infinity...for some reason I was thinking it was 0 times inifinty

Answer 10

and there is really no x inside the sine thing right?

Answer 11

and yes you are right @kayders1997

Answer 12

Thank you

Answer 13

I must have been thinking cos(pi/2)

Answer 14

cos(pi/2) is 0
and then your limit would be 0

Answer 15

No

Answer 16

if it was cos(pi/2) *x as x approaches infinity

Answer 17

Maybe idk

Answer 18

maybe?

Answer 19

what are you referring to with the maybe

Answer 20

My teacher said that if it was 0 and inifinty it could be 0 infinity or in between

Answer 21

\[\lim_{x \rightarrow \infty} c x=c \cdot \lim_{x \rightarrow \infty}x\]

Answer 22

where c is a constant

Answer 23

0 is a constant
you can pull it outside the limit

Answer 24

Okay

Answer 25

however if you had this
\[\lim_{x \rightarrow \infty} \frac{1}{x} \cdot x\]
you cannot do the following
\[\lim_{x \rightarrow \infty} \frac{1}{x} \cdot \lim_{x \rightarrow \infty} x\]
because one of the limits do not exist (that second one there)
and also this form is indeterminate 0*infty

Answer 26

but we know x/x=1

Answer 27

the so the limit here is 1

Answer 28

for this example

Answer 29

but you had a constant multiple which could be brought outside the limit always

Answer 30

Okay thank you