7. Computing Limits

b. When Limit Laws Don't Apply

Limits without Laws

1. Limit Tricks

d. Put Terms over a Common Denominator

This trick says to put all terms over a common denominator whether they appear in a numerator or denominator or not in a fraction at all. It is most appropriate for the indeterminate forms \(\dfrac{0}{0}\), \(\dfrac{\infty}{\infty}\), \(0\cdot\infty\) or \(\infty-\infty\).

Compute \(\displaystyle \lim_{x\to0}\left(\dfrac{1}{x^2}-\dfrac{1}{x^2+x^4}\right)\).

If we plug in \(x=0\), we see this has the indeterminate form \(\infty-\infty\). We put it over a common denominator: \[\begin{aligned} \lim_{x\to0}\left(\dfrac{1}{x^2}-\dfrac{1}{x^2+x^4}\right) &=\lim_{x\to0}\dfrac{(x^2+x^4)-x^2}{x^2(x^2+x^4)} \\ &=\lim_{x\to0}\dfrac{x^4}{x^4+x^6} \end{aligned}\] This now has the form \(\dfrac{0}{0}\). So we divide numerator and denominator by the largest term in the denominator which is \(x^4\). Why?

\[\begin{aligned} \lim_{x\to0}\left(\dfrac{1}{x^2}-\dfrac{1}{x^2+x^4}\right) &=\lim_{x\to0}\dfrac{x^4\dfrac{}{}}{(x^4+x^6)\dfrac{}{}} \,\dfrac{\;\dfrac{1}{x^4}\;}{\dfrac{1}{x^4}} \\ &=\lim_{x\to0}\dfrac{1}{1+x^2}=1 \end{aligned}\]

The following exercise is the same limit as in the last exercise on the page on dividing by the largest term in the denominator. This time you should put it over a common denominator.