# How do you use the limit definition to find the slope of the tangent line to the graph 3x^2-5x+2 at x=3?

Apr 10, 2016

Do a lot of algebra after applying the limit definition to find that the slope at $x = 3$ is $13$.

#### Explanation:

The limit definition of the derivative is:
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

If we evaluate this limit for $3 {x}^{2} - 5 x + 2$, we will get an expression for the derivative of this function. The derivative is simply the slope of the tangent line at a point; so evaluating the derivative at $x = 3$ will give us the slope of the tangent line at $x = 3$.

With that said, let's get started:
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{3 {\left(x + h\right)}^{2} - 5 \left(x + h\right) + 2 - \left(3 {x}^{2} - 5 x + 2\right)}{h}$
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{3 \left({x}^{2} + 2 h x + {h}^{2}\right) - 5 x - 5 h + 2 - 3 {x}^{2} + 5 x - 2}{h}$
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\cancel{3 {x}^{2}} + 6 h x + 3 {h}^{2} - \cancel{5 x} - 5 h + \cancel{2} - \cancel{3 {x}^{2}} + \cancel{5 x} - \cancel{2}}{h}$
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{6 h x + 3 {h}^{2} - 5 h}{h}$
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\cancel{h} \left(6 x + 3 h - 5\right)}{\cancel{h}}$
$f ' \left(x\right) = {\lim}_{h \to 0} 6 x + 3 h - 5$

Evaluating this limit at $h = 0$,
$f ' \left(x\right) = 6 x + 3 \left(0\right) - 5 = 6 x - 5$

Now that we have the derivative, we just need to plug in $x = 3$ to find the slope of the tangent line there:
$f ' \left(3\right) = 6 \left(3\right) - 5 = 18 - 5 = 13$

Apr 10, 2016

See the explanation section below if your teacher/textbook uses ${\lim}_{x \rightarrow a} \frac{f \left(x\right) - f \left(a\right)}{x - a}$

#### Explanation:

Some presentations of calculus use, for the defintion of the slope of the line tangent to the graph of $f \left(x\right)$ at the point where $x = a$ is ${\lim}_{x \rightarrow a} \frac{f \left(x\right) - f \left(a\right)}{x - a}$ provided that the limit exists.

(For example James Stewart's 8th edition Calculus p 106. On page 107, he gives the equivalent ${\lim}_{h \rightarrow 0} \frac{f \left(a + h\right) - f \left(a\right)}{h}$.)

With this definition, the slope of the tangent line to the graph of $f \left(x\right) = 3 {x}^{2} - 5 x + 2$ at the point where $x = 3$ is

${\lim}_{x \rightarrow 3} \frac{f \left(x\right) - f \left(3\right)}{x - 3} = {\lim}_{x \rightarrow 3} \frac{\left[3 {x}^{2} - 5 x + 2\right] - \left[3 {\left(3\right)}^{2} - 5 \left(3\right) + 2\right]}{x - 3}$

$= {\lim}_{x \rightarrow 3} \frac{3 {x}^{2} - 5 x + 2 - 27 + 15 - 2}{x - 3}$

$= {\lim}_{x \rightarrow 3} \frac{3 {x}^{2} - 5 x - 12}{x - 3}$

Note that this limit has indeterminate form $\frac{0}{0}$ because $3$ is a zero of the polynomial in the numerator.
Since $3$ is a zero, we know that $x - 3$ is a factor. So we can factor, reduce and try to evaluate again.

$= {\lim}_{x \rightarrow 3} \frac{\cancel{\left(x - 3\right)} \left(3 x + 4\right)}{\cancel{\left(x - 3\right)}}$

$= {\lim}_{x \rightarrow 3} \left(3 x + 4\right) = 3 \left(3\right) + 4 = 13$.
The limit is $13$, so the slope of the tangent line at $x = 3$ is $13$.