Trigonometric Limits

[rangnekarmitali]

I have a limit question that is the limit as x approaches 0 of tan(2x)/tan(pi x). I know that the limit exists I just don't know how to break it down. Are there rules that I'm forgetting?

 

[Deleted member 4993]

I have a limit question that is the limit as x approaches 0 of tan(2x)/tan(pi x). I know that the limit exists I just don't know how to break it down. Are there rules that I'm forgetting?

How do you know that? What is the value of that limit?

Please show us what you have tried and exactly where you are stuck.​

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[rangnekarmitali]

I have a limit question that is the limit as x approaches 0 of tan(2x)/tan(pi x). I know that the limit exists I just don't know how to break it down. Are there rules that I'm forgetting?

At first I thought the limit doesn't exist because I broke the limit down as (sin2x/cos2x)(cos pi x/sin pi x). I thought that since as x approaches zero, sinx approaches zero as well, which would make the limit approach infinity times sin2x/cos2x. When I went to check my answer in the textbook, it said that the limit is 2/pi. I have to solve this limit without L'Hopital's rule, but I don't know how to move on from here.

 

[Deleted member 4993]

At first I thought the limit doesn't exist because I broke the limit down as (sin2x/cos2x)(cos pi x/sin pi x). I thought that since as x approaches zero, sinx approaches zero as well, which would make the limit approach infinity times sin2x/cos2x. When I went to check my answer in the textbook, it said that the limit is 2/pi. I have to solve this limit without L'Hopital's rule, but I don't know how to move on from here.

Hint:

\(\displaystyle \lim_{a\to 0}\frac{tan(a)}{a} \ = 1 \)

Write the given Left_Hand_side as product of two functions - each resembling the LHS of the equation above.

 

[rangnekarmitali]

Hint:

\(\displaystyle \lim_{a\to 0}\frac{tan(a)}{a} \ = 1 \)

Write the given Left_Hand_side as product of two functions - each resembling the LHS of the equation above.

Oh ok, thank you!

 

[Deleted member 4993]

Oh ok, thank you!

If you get stuck on the way - do come back and show your work. We will unstuck you!