Help with limits

[jordan]

I've got a test tomorrow. As I was working on some of the review problems, I came across two I couldn't answer:

 

[math]\lim_{x \to 0} \frac{sin[(\pi/6)+\Delta{x}]-(1/2)}{\Delta{x}}[/math]

 

 

 

[math]\lim_{x \to +0} \frac{csc(2x)}{x}[/math]

 

 

 

Please help with an analytical proof. Oh, and how do I type the symbol for "all real numbers".

[Callipygous]

god i hate limits... sorry, this is the first thing i have struggled with in math... : P

[DerSpooky]

I've got a test tomorrow. As I was working on some of the review problems' date=' I came across two I couldn't answer:

 

-(1/2)}{Delta{x}}');"]

 

 

 

 

I didn't work through them all the way, but here are some ideas:

 

In the first problem, notice that as x approaches 0 (substitute x=0) you are left with [sin(pi/6)=1/2]-1/2, giving you the indeterminate form 0/0. Use L'Hopitals Rule.

 

The second one is a little involved, but only by identities. Keep in mind i havn't worked it out all the way but I think it will work. csc(2x) is equal to the root of 1+cot^2(2x). cot^2(2x)is equal to cos^2(2x)/sin^2(2x). sin^2(2x) is equal to 1-cos^2(2x). The numerator should now be in the form of the root of 1+cos^2(2x)/(1-cos^2(2x)). Multiply the top and bottom of the second term under the root by the conjugate base of the denominater (1+cos^2(2x)). When you substitute +0 for x the whole numerator should reduce down to the root of 1. You are the left with the lim 1/x.

 

hope this helps...

[jordan]

Thanks for the help DerSpooky. For the first one, I know you're getting a little too complex because we haven't learned about L'Hopitals Rule. On second thought, your method is probably simpler, but there has to be another way that doesn't use that rule. Since you mentioned it, though, what is this rule?

 

Also on the first one, I have managed to narrow it down to

 

[math]\frac{cos(\Delta{x})+\sqrt{3}-1}{2(\Delta{x})}[/math]

 

The second one sounds like it might work. I'll try it out it see what happens. Thanks again.

[DerSpooky]

(with a little refresher from a text book) L'Hopitals rule says that if you have two functions f and g that are differentiable over an interval (a,b) except maybe at a point c within the interval, and given that the lim(x to c) f(x)/g(x) is in the indeterminate form 0/0 or infinity/infinity, then if the lim (x to c) f'(x)/g'(x)=L (or +/- infinity) where the g'(x) doesn't equal 0 we can say

 

limf(x)/g(x)=limf'(x)/g'(x)

translation...when your limit goes to 0/0, or infinity/infinity, you can take the limit of the derivatives of both functions (don't use the quotient rule!) and you will get the same answer, as long as the derivative of the denominater doesn't equal zero. (this cuts out some of the finer details above that are still important!)

[jordan]

Unfortunatly, we haven't gotten to derivatives yet either.

[pulkit]

In the first one use the formula for Sin(a)-Sin(b) , you will end up with a term Sin(x) upon x which in the limit will turn out to be one. Note that youĺl need to put 1/2 = Sin (pi/6)