Evaluate the following limit or show it does not exist?

[LilTaru]

Homework Statement

lim x-->pi/6 sin(x + pi/3) - 1/(x - pi/6)

Homework Equations

The Attempt at a Solution

Everytime I try to solve it (i.e. use the identity for sin(a + b) it still turns out to not exist... I was just wondering if I was correct and it does not exist, or if there was some way to further simplify it?

 

[Mark44]

I don't see any need to use the identity for sin(a + b).
\lim_{x \to \pi/6} sin(x + \pi/3) - \frac{1}{x - \pi/6}
= \lim_{x \to \pi/6} sin(x + \pi/3) - \lim_{x \to \pi/6} \frac{1}{x - \pi/6}
= sin(\pi/2) - \lim_{x \to \pi/6} \frac{1}{x - \pi/6}

The first expression is just 1, so the limit in this problem hinges on whether the last limit exists.

 

[Petek]

I wonder if the problem is actually

\lim_{x \to \pi/6} \frac{sin(x + \pi/3) - 1}{x - \pi/6}

since this limit is more interesting.

 

[LilTaru]

Yes, Petek, you are correct! I have no idea how to solve it... It seems like I go in circles!

 

[jdc15]

Maybe our friend L'Hopital can help: http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule"

 

[LilTaru]

Seems to work, just want to check the result:

lim x-->pi/6 f¹(x)/g¹(x)
= lim x-->pi/6 cos(x + pi/3) - 1/(1)
= -1/1
= -1
? Is this correct?

 

[Mark44]

USE PARENTHESES!

You could have saved time by using parenthese in your limit expression. If you would have written this -
lim x-->pi/6 [sin(x + pi/3) - 1]/(x - pi/6)
I would have understood what you were trying to do.



The answer you got is not correct. One of your derivatives is incorrect.

 

[LilTaru]

Which one:

The derivative of x - pi/6 = 1 -0 = 1 Does it not because the derivative of x is 1 and the derivative of a constant is 0...

The derivative of sin(x + pi/3) - 1 = cos(x + pi/3) Since the derivative of sinx is cosx

Oh... which would make the limit = 0/1, which equals 0?

 

[Mark44]

Yes.

You had lim x-->pi/6 cos(x + pi/3) - 1/(1)
As you see, the numerator is incorrect.

BTW, write the fraction as (cos(x + pi/3) - 1)/1. IOW, put parentheses around the whole numerator. Since there is just one term in the denominator it doesn't need parentheses.

 

[LilTaru]

Thank you! Your help was very beneficial! Helped out a lot!

 

[joshiemen]

Yes.

You had lim x-->pi/6 cos(x + pi/3) - 1/(1)
As you see, the numerator is incorrect.

BTW, write the fraction as (cos(x + pi/3) - 1)/1. IOW, put parentheses around the whole numerator. Since there is just one term in the denominator it doesn't need parentheses.

Mark-is it possible to solve this without resorting to lohpital's rule? Just using limits?

 

[Mark44]

It might be, but nothing obvious comes to mind.

 

[joshiemen]

I wonder if the problem is actually

\lim_{x \to \pi/6} \frac{sin(x + \pi/3) - 1}{x - \pi/6}

since this limit is more interesting.

So, I tried solving, by replacing 1 first with sin(pi/2)

so it, looks

_{(sin(x+pi/3)-sin(pi/2))/(x-pi/6)}

next applied product to sum formulas:

looks like this:

_{2*cos((x+pi/3+pi/2)/2)*sin(((x+pi/3-pi/2)/2)/((x-pi/6))}

skipping simplifying steps here, it left me with cos(pi/2) which is clearly...0...

does this make sense? Can anyone write it up nicely in latex or complete...solution much appreciated...on a piece of paper I got it...cannot type with latex for some reason...

 

[Bohrok]

is it possible to solve this without resorting to lohpital's rule? Just using limits?

Yes

Use the substitution u = x - \pi/6

As x → \pi/6, u → 0. Also, x = u + \pi/6

Then sin(x + \pi/3) = sin(u + \pi/6 + \pi/3) = sin(u + \pi/2) = cos u

Finally,
\lim_{x \to \pi/6} \frac{\sin(x + \pi/3) - 1}{x - \pi/6} \Longrightarrow \lim_{u \to 0} \frac{\cos u - 1}{u} = 0A good thing to try for limits like this is substituting u = the denominator, then you can usually rewrite it into a well-known limit that you can easily evaluate.

 

[joshiemen]

Yes

Use the substitution u = x - \pi/6

As x → \pi/6, u → 0. Also, x = u + \pi/6

Then sin(x + \pi/3) = sin(u + \pi/6 + \pi/3) = sin(u + \pi/2) = cos u

Finally,
\lim_{x \to \pi/6} \frac{\sin(x + \pi/3) - 1}{x - \pi/6} \Longrightarrow \lim_{u \to 0} \frac{\cos u - 1}{u} = 0A good thing to try for limits like this is substituting u = the denominator, then you can usually rewrite it into a well-known limit that you can easily evaluate.

wow, this is way simpler and beautiful than how I solved it! THANK you!