Question

How do you use the limit definition to find the slope of the tangent line to the graph `f(x) = (x-1)^3` at x=2?

Answer 1

The slope is `3`. Please see the explanation section, below.

Explanation:

Although we could use the defintion in the form `lim_(hrarr0)(f(x+h)-f(x))/h`,

I think that the algebra is simpler if we use the equivalent

The slope of the tangent line to the graph of `f(x)` at the point where `x=2` is

`lim_(xrarr2)(f(x)-f(2))/(x-2)`.

In this case, we get:

`lim_(xrarr2)((x-1)^3-(2-1)^3)/(x-2)`.

Obviously (as expected), the initial form of this limit is indeterminate, so we'll need to reduce the fraction and try again.

Expanding `(x-1)^3` either by multiplying or by the binomial expansion, and then simplifying, we get

`lim_(xrarr2)((x^3-3x^2+3x-1)-(1))/(x-2) = lim_(xrarr2)(x^3-3x^2+3x-2)/(x-2)`

Because `2` is a zero of the polynomial in the numerator, we know that `x-2` is a factor. By division, or by trial and error or grouping, we get

`x^3-3x^2+3x-2 = x^3-2x^2-x^2+2x+x-2`

`= (x-2)(x^2-x+1)`.

So our limit becomes:

`lim_(xrarr2)((x-1)^3-(2-1)^3)/(x-2) = lim_(xrarr2)((x^3-3x^2+3x-1)-(1))/(x-2)`

`= lim_(xrarr2)(x^3-3x^2+3x-2)/(x-2)`

`= lim_(xrarr2)(cancel((x-2))(x^2-x+1))/cancel((x-2))`

`= 4-2+1`

`=3`