Question

Help! Lim as x approaches pi/6 sinx-1/2/(x-pi/6)

Answer 1

\[\lim_{x \rightarrow \pi/6} \frac{ sinx-1/2 }{ x-\pi/6 }\]

Answer 2

use L'Hopitals rule since you have an indeterminate form of 0/0

Answer 3

Doubt she knows l'hopitals.

Answer 4

Do I start out with sinx-1/2 * sinx+1/2?

Answer 5

ahhh, basically take derivative of top and derivative of bottom and take that limit

Answer 6

Yeah, but assuming shes not supposed to know l'hopitals, we dont want to just teach it, shes supposed to find it a different way.

Answer 7

using l hospitals rule,
cosx-0/1
now put the limit value in place of x.
cos pie/6
root3/2

Answer 8

cos x/1 then it's just cos pi/6 which is sqrt3/2

Answer 9

I dont think you can take the limit without it though

Answer 10

we nicknamed lhopitals the limit breaker lol

Answer 11

Oh, we didn't learn that yet.

Answer 12

this might be the problem to get you there

Answer 13

I have a suggestion then, thats not l'hopitals. Doesnt mean itll work, but....

Answer 14

u havent learned l hospitals?then first go through the theory part in ur text book.

Answer 15

its pretty easy, if you get a form of 0/0 or infinite/infinite you can use lhopitals rule

Answer 16

if you take the limit of that then that's the limit for the function

Answer 17

Do I just times the bottom and top by sinx+1/2? Will that work?

Answer 18

that might you can always multiply a function by 1 in any form

Answer 19

but it really just makes life hard lol

Answer 20

Well, that's the way the professor wants it lol.

Answer 21

in l hospitals rule,first u have to check weather the numerator and denominator both are coming to zero by putting the value of limit in place of x.

Answer 22

Of course the denominator has a zero.

Answer 23

yea if that's the case then manipulate the equation to a form that you can plug in the limit for without dividing by 0,

Answer 24

which class?

Answer 25

Calculus I

Answer 26

ahhh yea you learn L'hopitals in Calc 2

Answer 27

you need it to break open harder limits

Answer 28

The other way is to put the function in your calculator and get ever so close to pi/6 but not at pi/6

Answer 29

We have to prove it though by writing it out.

Answer 30

basically however you do it the answer is sqrt3/2 so if you get that you're good. =)

Answer 31

Actually, my question is how do you get to that answer.. but thanks lol

Answer 32

No, it wont. You HAVE to get rid of the x-pi/6 deal, that won't go away. Like I said, a suggestion that may or may not work.

So, substitution time.

\[u = x-\frac{ \pi }{ 6 }\]
therefore
\[x = u+ \frac{ \pi }{ 6 }\]
\[\frac{ \sin(u+\frac{ \pi }{ 6 })-\frac{ 1 }{ 2 } }{ u }\] Use sum of sines formula to expand the sin portion:

\[\frac{ sinucos(\frac{ \pi }{ 6 })+\sin(\frac{ \pi }{ 6 })cosu-\frac{ 1 }{ 2 } }{ u }\]
\[\frac{ \frac{ \sqrt{3} }{ 2 }sinu+\frac{ 1 }{ 2 }cosu-\frac{ 1 }{ 2 } }{ u }\]
Now because of theu-substitution, as x goes to pi/6, u goes to 0. So we have to keep this in mind as I finish. Now I'll split this into two fractions:

\[\lim_{u \rightarrow 0}\frac{ \frac{ \sqrt{3} }{ 2 }sinu }{ u }+\frac{ \frac{ 1 }{ 2 }(cosu-1) }{ u }\]

Answer 33

Now there are two limit identities that can be used when the limit approaches 0. The first is
\[\lim_{x \rightarrow 0}\frac{ sinx }{ x }= 1\]
the second is
\[\lim_{x \rightarrow 0}\frac{ cosx -1 }{ x }= 0\]Well, looks like to me we have both of those identities in our two fractions. So as u goes to o, that left fraction that has sinu/u goes to 1. And the right fraction that has (cosu-1)/u becomes 0. SO that means you're left finally with
\[\lim_{u \rightarrow 0}\frac{ \sqrt{3} }{ 2 }(1) + \frac{ 1 }{ 2 }(0) = \frac{ \sqrt{3} }{ 2 }\]

Answer 34

Thank you so much! I actually understand it now. I forgot that I had to make u=x-a and x=u+a as x approaches a and u approaches 0. You get best response!

Answer 35

Its a really sneaky substitution. At first glance, the calc 2 method just looks like it may be the only way, but you can sometimes manipulate things in some awesome ways. Glad you understand what I did ^_^

Answer 36

Could you help me with one more thing by the way? How would I explain my answer for sin (.53)-1/2 over .53 - pi/6 is .86442 without using a calculator?

Answer 37

They gave me a hint that said pi/6 = .523598

Answer 38

It looks like the exact same problem.

Answer 39

Well nevermind, you helped enough lol. It says multiple choice (do not use calculator) then put the problem up and it says explain your answer. I don't know what's with that.

Answer 40

Well, you just would rewrite it into the same problem you had before. You just would let .53 be x and say its a limit as x goes to .53, then do the same thing.

Answer 41

Oh okay! That makes sense lol thank you.

Answer 42

yeah, sure xD