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:

\[ \lim_{x \to 0} \frac{sin(5x)} {x} \]

and

\[ \lim_{x \to \infty} \frac{sin(5x)} {x} \]

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 \(x \to \infty\).

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:

\[ \lim_{x \to 0} \frac{sin(x)} {x} = 1 \]

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

\[ \lim_{x \to 0} cos(x) = 1 \]

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

\[ \lim_{x \to 0} \frac{tan(x)} {x} = 1 \]

The large case

As we all know:

\[ -1 \le sin(x) \le 1 \]

So the limit is:

\[ \lim_{x \to \infty} \frac{sin(x)} {x} = 0 \]

Derivatives Involving Trig Functions

Using the definition of derivative:

\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)} {h} \]

and the conclusion above:

\[ \lim_{h \to 0} \frac{sin(h)} {h} = 1 \]

we can get:

\[ \begin{align} sin'(x) &= \lim_{h \to 0} \frac{sin(x + h) - sin(x)} {h} \\ &= \lim_{h \to 0} \frac {sin(x)cos(h) + cos(x)sin(h) - sin(x)} {h} \\ &= \lim_{h \to 0} (sin(x)(\frac{cos(h) - 1} {h}) + cos(x)(\frac{sin(h)}{h})) \\ &= \lim_{h \to 0} (sin(x) \times 0 + cos(x) \times 1) \\ &= cos(x) \end{align} \]

Use the samme method, we can get other equations:

\[ \begin{align} cos'(x) &= -sin(x) \\ tan'(x) &= \frac{1}{cos^2(x)} = sec^2(x) \\ sec'(x) &= sec(x)tan(x) \\ csc'(x) &= -csc(x)cot(x) \\ cot'(x) &= -csc^2(x) \end{align} \]