Trig Limits and Derivatives

So far, most of our limits and derivativess have involved only polynomials or poly-type functions. Lets expand our horizons by looking aat trig functions.

Limits Involving Trig Functions

Consider the following two limits:

$$\underset{x\to 0}{lim}\frac{sin(5x)}{x}$$

and

$$\underset{x\to \mathrm{\infty }}{lim}\frac{sin(5x)}{x}$$

They look almost the same, the only difference is that the first limit is taken as $x\to 0$ while the second it taken as $x\to \mathrm{\infty }$.

The small case

We know that $sin(0)=0$, so what does $sin(x)$ look like when $x$ is near $0$? It turns out that $sin(x)$ is approximately the same as $x$ itself:

The graphs of $y=sin(x)$ and $y=x$ are very similar, especially when $x$ is close to $0$. So we have the conclusion:

$$\underset{x\to 0}{lim}\frac{sin(x)}{x}=1$$

How about $cos(x)$? Well, $cos(0)=1$, so we write:

$$\underset{x\to 0}{lim}cos(x)=1$$

As for $tan(x)$, we can write $tan(x)$ as $\frac{sin(x)}{cos(x)}$ and get the equation:

$$\underset{x\to 0}{lim}\frac{tan(x)}{x}=1$$

The large case

As we all know:

$$-1\le sin(x)\le 1$$

So the limit is:

$$\underset{x\to \mathrm{\infty }}{lim}\frac{sin(x)}{x}=0$$

Derivatives Involving Trig Functions

Using the definition of derivative:

$$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$$

and the conclusion above:

$$\underset{h\to 0}{lim}\frac{sin(h)}{h}=1$$

we can get:

$$\begin{array}{rl}si{n}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{sin(x+h)-sin(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{sin(x)cos(h)+cos(x)sin(h)-sin(x)}{h}\\ & =\underset{h\to 0}{lim}(sin(x)(\frac{cos(h)-1}{h})+cos(x)(\frac{sin(h)}{h}))\\ & =\underset{h\to 0}{lim}(sin(x)\times 0+cos(x)\times 1)\\ & =cos(x)\end{array}$$

Use the samme method, we can get other equations:

$$\begin{array}{rl}co{s}^{\prime }(x)& =-sin(x)\\ ta{n}^{\prime }(x)& =\frac{1}{co{s}^2(x)}=se{c}^2(x)\\ se{c}^{\prime }(x)& =sec(x)tan(x)\\ cs{c}^{\prime }(x)& =-csc(x)cot(x)\\ co{t}^{\prime }(x)& =-cs{c}^2(x)\end{array}$$