# 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?

##### 2 Answers

Do a lot of algebra after applying the limit definition to find that the slope at

#### Explanation:

The limit definition of the derivative is:

If we evaluate this limit for *derivative* of this function. The derivative is simply the slope of the tangent line at a point; so evaluating the derivative at

With that said, let's get started:

Evaluating this limit at

Now that we have the derivative, we just need to plug in

See the explanation section below if your teacher/textbook uses

#### Explanation:

Some presentations of calculus use, for the defintion of the slope of the line tangent to the graph of

(For example James Stewart's 8th edition *Calculus* p 106. On page 107, he gives the equivalent

With this definition, the slope of the tangent line to the graph of

# = lim_(xrarr3)(3x^2-5x+2-27+15-2)/(x-3)#

# = lim_(xrarr3)(3x^2-5x-12)/(x-3)#

Note that this limit has indeterminate form

Since

# = lim_(xrarr3)(cancel((x-3))(3x+4))/cancel((x-3))#

# = lim_(xrarr3)(3x+4) = 3(3)+4 = 13# .

The limit is#13# , so the slope of the tangent line at#x=3# is#13# .