7. Computing Limits

b. When Limit Laws Don't Apply

Limits without Laws

1. Limit Tricks

b. Expand and Cancel

This trick says to expand the numerator and the denominator and then cancel any common factors. It is most appropriate for the indeterminate forms \(\dfrac{0}{0}\).

Compute \(\displaystyle \lim_{x\to0}\dfrac{(x-2)^2-4}{x}\).

If we plug \(x=0\) into the numerator and denominator, we see this has the indeterminate form \(\dfrac{0}{0}\). To simplify this, we expand and symplify the numerator and cancel: \[\begin{aligned} \lim_{x\to0}&\dfrac{(x-2)^2-4}{x} =\lim_{x\to0}\dfrac{(x^2-4x+4)-4}{x} \\ &=\lim_{x\to0}\dfrac{x^2-4x}{x} =\lim_{x\to0}(x-4)=-4 \end{aligned}\]

Sometimes we need to use \(x\) as a parameter (a constant) in the limit and then we use \(h\) as the variable in the limit.

Compute \(\displaystyle \lim_{h\to0}\dfrac{(x+h)^3-3(x+h)^2-x^3+3x^2}{h}\).

If we plug \(h=0\) into the numerator and denominator, we see this has the indeterminate form \(\dfrac{0}{0}\) for all \(x\). To simplify this, we expand and symplify the numerator and then cancel the \(h\) from the denominator: \[\begin{aligned} \lim_{h\to0}&\dfrac{(x+h)^3-3(x+h)^2-x^3+3x^2}{h} \\ &=\lim_{h\to0}\dfrac{(x^3+3x^2h+3xh^2+h^3)-3(x^2+2xh+h^2)-x^3+3x^2}{h} \\ &=\lim_{h\to0}\dfrac{3x^2h+3xh^2+h^3-6xh-3h^2}{h} \\ &=\lim_{h\to0}(3x^2+3xh+h^2-6x-3h) =3x^2-6x \end{aligned}\]