Question

evaluate limit as x approaches zero from the right. sin (x) ln (3x).

Answer 1

Have you learned the Squeeze/Sandwhich theorem?

Answer 2

Yeah

Answer 3

It's not the whole probelm that i need help with it's just the part at the very end where you have sin (x)/x = 1

Answer 4

Wait so is your question with sin(x) ln(3x) or sin(x) / x?

Answer 5

it's dealing with that problem, just the very end of it. i 'm just wondering how the limit as x approaches 0 from the right of sin(x)/x is 1

Answer 6

Its a Squeeze Theorem proof and its a geometrical one:

http://www.youtube.com/watch?v=Ve99biD1KtA

Answer 7

to understand that you need to know l'hopitals. Thats why in calc 1 they call these "Special limits"

Answer 8

ok i get you so far...

Answer 9

if you have

the form

\[\frac{0}{0}\], you can find the limit by take the derivative of the top and the derivative of the bottom and evaluate the limit again until you do not get a indeterminate form

Answer 10

so taking the derivative you get
\[\frac {cos(x)}{1}\]

as x goes to 0, this equals 1

Answer 11

that is why sinx/x equals 1

Answer 12

Oooo! ok thanks so much! One more question. How does the limit as x approaches 0 of cot (2x) equal infinity?

Answer 13

cot(2x) is 1/tan(2x)

as x->0, tan(2x) is going to tan(0) which is 0. so you have 1/0

Answer 14

this you cannot use the l'hopitals quite yet

one way i cant think of is using double angle stuff

Answer 15

1/0 is undefined not infinity

Answer 16

\[\frac{cos(2x)}{sin(2x)}\]

Answer 17

That's what I thought too but im doing homework on this online thing and they said that 0 was wrong and it was infinity so I j

Answer 18

anted to know why

Answer 19

no what i'm saying is that you cannot assume that it is infinity by seeing that it's undefined... there is a chance it is infinity though

Answer 20

okey doke

Answer 21

in this case it is infinity