# Thread: Problem trying to find limit of trigonometric function

1. ## Problem trying to find limit of trigonometric function

Hi,

I'm given the problem of finding the limit as x approaches 0 of tan3x/tan5x. To start, I replace tan with sin/cos, so that I get sin3x/cos3x/sin5x/cos5x. My professor wrote it out as sin3x/1 1/sin5x cos5x1/cos3x. He then somehow found a way to get a 3x to the denominator of the first term after factoring out a 3/5, and add 5x to the numerator of the second term. I'm not sure how he managed that, could someone help me to understand?

Thanks!

Kevin

2. ## Re: Problem trying to find limit of trigonometric function

I think I got it: he added 5x to the denominator of the first term and to the numerator of the second term, which both cancel out so is the same as multiplying by 1. He then multiplied the first term by 3/3, factored out 3/5, and was left with an easily solvable limit?

3. ## Re: Problem trying to find limit of trigonometric function

Originally Posted by KevinShaughnessy
Hi,

I'm given the problem of finding the limit as x approaches 0 of tan3x/tan5x. To start, I replace tan with sin/cos, so that I get sin3x/cos3x/sin5x/cos5x. My professor wrote it out as sin3x/1 1/sin5x cos5x1/cos3x. He then somehow found a way to get a 3x to the denominator of the first term after factoring out a 3/5, and add 5x to the numerator of the second term. I'm not sure how he managed that, could someone help me to understand?

Thanks!

Kevin
First, you need to know that \displaystyle \displaystyle \begin{align*} \lim_{x \to 0}\frac{\tan{(x)}}{x} = 1 \end{align*}. This should be obvious if you already know the standard limit \displaystyle \displaystyle \begin{align*} \lim_{x \to 0}\frac{\sin{(x)}}{x} = 1 \end{align*}. Then

\displaystyle \displaystyle \begin{align*} \lim_{x \to 0} \frac{\tan{(3x)}}{\tan{(5x)}} &= \lim_{x \to 0} \left[ \frac{15x}{15x} \cdot \frac{\tan{(3x)}}{\tan{(5x)}} \right] \\ &= \frac{3}{5} \lim_{x \to 0} \left[ \frac{\tan{(3x)}}{3x} \cdot \frac{5x}{\tan{(5x)}} \right] \end{align*}

I'm sure you can go from here.