Question

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?

Answer 1

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'(x)=lim_(h->0)(f(x+h)-f(x))/h`

If we evaluate this limit for `3x^2-5x+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'(x)=lim_(h->0)(3(x+h)^2-5(x+h)+2-(3x^2-5x+2))/h`
`f'(x)=lim_(h->0)(3(x^2+2hx+h^2)-5x-5h+2-3x^2+5x-2)/h`
`f'(x)=lim_(h->0)(cancel(3x^2)+6hx+3h^2-cancel(5x)-5h+cancel(2)-cancel(3x^2)+cancel(5x)-cancel(2))/h`
`f'(x)=lim_(h->0)(6hx+3h^2-5h)/h`
`f'(x)=lim_(h->0)(cancel(h)(6x+3h-5))/cancel(h)`
`f'(x)=lim_(h->0)6x+3h-5`

Evaluating this limit at `h=0`,
`f'(x)=6x+3(0)-5=6x-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'(3)=6(3)-5=18-5=13`

Answer 2

See the explanation section below if your teacher/textbook uses `lim_(xrarra)(f(x)-f(a))/(x-a)`

Explanation:

Some presentations of calculus use, for the defintion of the slope of the line tangent to the graph of `f(x)` at the point where `x=a` is `lim_(xrarra)(f(x)-f(a))/(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_(hrarr0)(f(a+h)-f(a))/h`.)

With this definition, the slope of the tangent line to the graph of `f(x) = 3x^2-5x+2` at the point where `x=3` is

`lim_(xrarr3)(f(x)-f(3))/(x-3) = lim_(xrarr3)([3x^2-5x+2]-[3(3)^2-5(3)+2])/(x-3)`

`= 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 `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_(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`.