How to Find the Limit of a Trigonometric Function

This Article will be using, as an example, a Trigonometric problem in other to demonstrate the process of finding the Limit of a Trigonometric function.

Things You'll Need

• Paper and
• Pencil.

Instructions

1. 

   To find the Limit of the Trigonometric function f(x) = (sinx)/x as x approaches Zero (0), we apply the "Squeeze Theorem". That is the Limit of (sinx)/x as x approaches (0),is (1). Please click on the Image for a better understanding.

2. 

   The Problem we will be working with in this article, is...Find the Limit, as x->0, of (sin2x)/(sin3x). To solve this we need to apply the Squeeze Theorem. Which is the Limit of (sinx)/(x) as x->0 = 1. Please click on the image for a better understanding.

3. 

   In order to find the limit of this function, we need to rewrite the initial function so that it has the form (sinx)/(x). To do this, we substitute the variables u and v for 2x and 3x respectively, as the arguements for the trigonometric expressions. That is, instead of (sin2x)/(sin3x) we will have [(sinu)/u]/[v/(sinv)], where 2x=u and 3x=v. When they are separated, the function becomes [(sinu)/(u)][(v)/(sinv)]. We need to mulitply the function by [u/v]. Please click on the image for a better understanding.

4. 

   Now, the Limit, as x -> 0, of [sin2x]/[sin3x] is equal to the Limit of (sin2x)/(2x) * (3x)/(sin3x) * (2/3), where * stands for multiply. Applying the Squeeze Theorem, we can set (sin2x)/(2x) and (3x)/(sin3x) both equal to 1. This will leave us with (1)*(1)*(2/3), which is equal to (2/3). Please click on the image for a better understanding.