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# Derivative as a limit

*Derivative* is the slope of the tangent line.

But what *is* the tangent line? (Silly answer: Line the slope of which is equal to the derivative).

What do we need it for? Recall the examples:

- Velocity
- Headlights
- Sling

i.e. it is used to study motion.

To find the tangent line, we only need its *slope*. We define its slope as the *limit of slopes* of the secant lines, lines determined by two points on the graph. For each $x \neq a$, draw a line through $(a,f(a))$ and $(x,f(x))$. We have infinitely many secant lines. Taske one:
$$\text{Slope of this line } = \frac{\text{rise}}{\text{run}} = \frac{f(x) - f(a)}{x-a}$$
Then the slope of the tangent line
$$\lim_{x \to a} \frac{f(x) - f(a)}{x-a}$$

So, as we move $x$ toward $a$, the secant lines turn and approach the tangent line, provided the limit exists. This number is called *the derivative of $f$ at $x = a$*. Note: this is not limit of $f$ but of the *"difference quotient"*, a new function made from $f$.

**Review exercise.** Compute and justify:
$$\lim_{x\to \infty} \frac{-x^{2} + 2x - 7}{2x^{2} + 77}$$

**Example.** Find the tangent line to $y = x^{2}$ at $(1,1)$.

Slope = the derivative of $y=x^{2}$ at $x=1$.

Use $f(x) = x^{2}$, $a =1$ in the definition. $$\begin{aligned} f^{\prime} &= \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \\ & = \lim_{x \to 1} \frac{x^{2} - 1^{2}}{x - 1} \\ & = \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} \\ & = \lim_{x \to 1} ( x + 1 ) \\ & = 1 + 1 \\ &= 2 \quad \text{ Slope } \end{aligned}$$

The point-slope form of the line: $$ y - 1 = 2(x - 1)$$ Consider: $$f^{\prime}(a) = \lim_{x \to a} \frac{\overbrace{f(x) - f(a)}^{?}}{\underbrace{x - a}_{0}}$$ What are the limits of the numerator and the denominator? $$\lim_{x \to a} \underbrace{\left(f(x) - f(a)\right)}_{\text{If continuous, substitute}} = f(a) - f(a) = 0$$ This is an indeterminate expression. So, possibly the derivative exists.

What if $f$ is *not* continuous at $x = a$? Then
$$\lim_{x \to a} \frac{\overbrace{f(x) - f(a)}^{\text{This limit does not exist.}}}{\underbrace{x - a}_{0}}$$
Then $f^{\prime}(a)$ does not exist.

Therefore...

**Theorem.** If $f$ is not continuous at $x = a$, then the derivative $f^{\prime}(a)$ does not exist.

We say that $f$ isn't *differentiable*.

**Example.** Consider
$$f(x) = x^{2} \to f^{\prime}(1) = \lim_{x \to a} \frac{x^{2} - 1^{2}}{x-1}$$
To resolve $\frac{0}{0}$, we factor numerator. It's easy.

Factoring, in general, is hard...

**Alternative definition of derivative**

Rename $h = x - 1$. Then

$$f^{\prime}(a) = \lim_{x \to a} \frac{f(x) - f(a)}{\underbrace{x - a}_{h = x - a}}$$ Since $h = x - a$, then $x = a + h$

Rewrite the definition: $$ f^{\prime}(a) = \lim_{x \to a} \frac{f(a + h) - f(a)}{h} $$

This makes life easier...

**Example.** Given
$$f(x) = x^{2}.$$

$$\begin{aligned} f^{\prime}(1) &= \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} \\ & = \lim_{h \to 0} \frac{(1 + h)^{2} - 1^{2}}{h} \\ & = \lim_{h \to 0} \frac{1^{2} + 2h + h - 1}{h} \\ & = \lim_{h \to 0} \frac{2h + h^{2}}{h} \qquad \frac{0}{0} \\ & = \lim_{h \to 0} (2 + h^{2}) = 2 + 0^{2} = 2 \end{aligned}$$

**Problem.** One can estimate the derivative from the graph. Let's find the derivative at the red point.

$$\text{Derivative } = \text{ Slope of the tangent line}$$

- Find it. It's the green line.
- Find its slope. $\frac{\text{Rise}}{\text{Run}} = \frac{14}{3} \approx 5$.

To better estimate the slope, draw a triangle as large as possible.

Here we get algebra from the graph. It's typically vice versa.

Find $f^{\prime}(1)$ for $f(x) = \frac{1}{x}$.

$$\begin{aligned} f^{\prime} &= \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} \qquad \to \frac{0}{0} \\ &= \lim_{h \to 0} \frac{\frac{1}{1+h} - \frac{1}{1}}{h} \qquad \to \frac{0}{0} \\ &= \lim_{h \to 0} \frac{\frac{1}{1+h} - \frac{1+h}{1+h}}{h} \\ &= \lim_{h \to 0} \frac{1 - (1-h)}{1+h} \cdot \frac{1}{h} \\ &= \lim_{h \to 0} \frac{-h}{1 + h}\cdot \frac{1}{h} \\ &= \lim_{h \to 0} -\frac{1}{1 + h} \\ &= -\frac{1}{1+0} = -1 \end{aligned}$$ This is a rational function, continuous on its domain. To evaluate the limit, plug in.

**Example.** Evaluate:
$$\lim_{x \to 5} \frac{2^{x} - 32}{x - 5}. $$
It's just a limit. But we recognize that this is the derivative of some function. Compare to the definition
$$ f^{\prime}(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} $$
and match.

So $$\begin{gathered} a = 5 \\ f(x) = 2^{x} \\ f(5) = 2^{5} = 32 \end{gathered}$$

So our limit, $f^{\prime}(5)$ for $f(x) = 2^{x}$. Compute: $$f^{\prime}(x) = (2^{x})^{\prime} = 2^{x} \ln 2, $$ so $$f^{\prime}(5) = 2^{5} \ln 2 = 32 \ln 2.$$