Limits of Trig Functions

There are two important limits involving trig functions that will be used later when finding the derivative of trig functions here. We can use the Squeeze Theorem to evaluate both of them. First,

$\lim_{h\to0}\frac{\sin h}{h} = 1$.

Try proving

$\lim_{h\to0}\frac{\cos h - 1}{h} = 0$

similarly!

(Videos) Limit Problems with Trig

by PatrickJMT

Here's a video by patrickJMT showing some problems of limits involving trig functions. Below are some more examples where he does some more complicated limits.

(Video) Calculating a Limit Involving $\frac{\sin x}{x}$ as $x$ Approaches Zero

by PatrickJMT

Here's a video by patrickJMT showing you how to find the limit of some trig functions.

There's a very helpful limit to know when solving these problems. It's worth memorizing this. $$\lim_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1=\lim_{\theta\rightarrow0}\frac{\theta}{\sin\theta}$$ There is a way to solve this limit to prove that it equals 1 but for now just take it at face value. Once we learn about the squeeze theorem or l'Hospital's rule we can solve this.

He then starts solving some examples using this identity. $$\lim_{x\rightarrow0}\frac{\sin3x}{x}$$ The idea behind this is whatever we have next to the sine we want to have the same thing in the denominator. In this example we have a $3x$ in the numerator so we want to have a $3x$ in the denominator. $$\begin{align} &\lim_{x\rightarrow0}\frac{\sin3x}{x}\color{red}{\cdot\frac33}\\ &=\lim_{x\rightarrow0}\frac{3\cdot\sin3x}{3x}\\ &=\lim_{x\rightarrow0}3\cdot\frac{\sin3x}{3x}\\ &=\lim_{x\rightarrow0}3\cdot1=3 \end{align}$$ So $$\lim_{x\rightarrow0}\frac{\sin3x}{x}=3$$

The next example he solves is a little more complicated. $$\lim_{t\rightarrow0}\frac{\tan6t}{\sin2t}$$ We can begin by first separating the functions. $$=\lim_{t\rightarrow0}\frac{\tan6t}1\cdot\frac1{\sin2t}$$ Next we can rewrite the tangent function and then separate. $$\begin{align} &=\lim_{t\rightarrow0}\frac{\sin6t}{\cos6t}\cdot\frac{1}{\sin2t}\\ &=\lim_{t\rightarrow0}\frac{\sin6t}{}\cdot\frac{1}{\cos6t}\cdot\frac{}{\sin2t}\\ \end{align}$$ Now that we have isolated sine terms we can start to turn them into the form our identity takes above. In the first part we have $\sin6t$ so we want to put a $6t$ on the bottom. $$=\lim_{t\rightarrow0}\frac{\sin6t}{6t}\cdot\frac{1}{\cos6t}\cdot\frac{}{\sin2t}$$ We still need to multiply a $6t$ into the numerator in order to make it equal so keep that in mind. Next we look at the $\frac{1}{\sin2t}$. We need a $2t$ in the numerator in order to match our identity. Well we already have a $t$ from the $\sin6t$ so that takes care of that. We can now multiply the 6 on the end. $$=\lim_{t\rightarrow0}\frac{\sin6t}{6t}\cdot\frac{1}{\cos6t}\cdot\frac{t}{\sin2t}\cdot\frac{6}{}$$ We need a $2$ in the numerator of the second sine term in order for our identity to work. So we can put one in the numerator and attach on a $2$ in the denominator to make it equal. $$=\lim_{t\rightarrow0}\frac{\sin6t}{6t}\cdot\frac{1}{\cos6t}\cdot\frac{2t}{\sin2t}\cdot\frac62\\ =1\cdot\frac11\cdot1\cdot3=3$$ So $$\lim_{t\rightarrow0}\frac{\tan6t}{\sin2t}=3$$