7. Computing Limits

b. When Limit Laws Don't Apply

Limits without Laws

1. Limit Tricks

c. Divide by the Largest Term in the Denominator

This trick says to identify the largest term in the denominator and then to divide the numerator and the denominator by that term. As we identify the largest term and divide by it, we can ignore any numerical coefficients. It is most appropriate for the indeterminate forms \(\dfrac{0}{0}\) or \(\dfrac{\infty}{\infty}\) when the numerator and denominator are generalized polynomials or even more general functions.

Compute \(\displaystyle \lim_{x\to0^+}\dfrac{6x^4-3x^2}{5x^4+7x^2}\).

If we plug \(x=0\) into the numerator and denominator, we see this has the indeterminate form \(\dfrac{0}{0}\). To simplify this, we first need to identify the largest term in the denominator. To do this we don't really need to look at the coefficients \(5\) and \(7\). For \(x\) close to \(0\), which is bigger: \(x^4\) or \(x^2\)? If \(x=\dfrac{1}{2}\), then \(x^4=\dfrac{1}{16}\) and \(x^2=\dfrac{1}{4}\). So \(x^2\) is bigger. So we divide numerator and denominator by \(x^2\): \[\begin{aligned} \lim_{x\to0^+}\dfrac{6x^4-3x^2}{5x^4+7x^2} &=\lim_{x\to0^+}\dfrac{(6x^4-3x^2)\dfrac{}{}}{(5x^4+7x^2)} \,\dfrac{\;\dfrac{1}{x^2}\;}{\dfrac{1}{x^2}} \\ &=\lim_{x\to0^+}\dfrac{6x^2-3}{5x^2+7} =-\,\dfrac{3}{7} \end{aligned}\]

So why do we divide by the largest term in the denominator, rather than the numerator? Because then that largest term becomes a constant and the other terms are smaller. If we divided by the largest term in the numerator, then the denominator might not become non-zero and we might still not be able to compute the limit. In the example above, the \(7x^2\) became \(7\), a constant and the \(5x^4\) became \(5x^2\) which goes to \(0\) as \(x\) goes to \(0\). The denominator is now non-zero and we can apply the ordinary Quotient Law.

Compute \(\displaystyle \lim_{x\to0^+}\dfrac{6x^4-3x}{5x^4+7x^2}\). (Comparing to the previous example, notice the different power for the term with a coefficient of \(3\).)

The largest term in the denominator is still \(x^2\). So we divide numerator and denominator by \(x^2\): \[\begin{aligned} \lim_{x\to0^+}\dfrac{6x^4-3x}{5x^4+7x^2} &=\lim_{x\to0^+}\dfrac{(6x^4-3x)\dfrac{}{}}{(5x^4+7x^2)} \,\dfrac{\;\dfrac{1}{x^2}\;}{\dfrac{1}{x^2}} \\ &=\lim_{x\to0^+}\dfrac{6x^2-\dfrac{3}{x}}{5x^2+7} =-\infty \end{aligned}\] In the last step, the denominator is still non-zero, \(7\), but now the \(\dfrac{3}{x}\) term in the numerator goes to \(\infty\).

Compute \(\displaystyle \lim_{x\to0^+}\dfrac{6x^4-3x^3}{5x^4+7x^2}\). (Notice the different power for the term with a coefficient of \(3\).)

The largest term in the denominator is still \(x^2\). So we divide numerator and denominator by \(x^2\): \[\begin{aligned} \lim_{x\to0^+}\dfrac{6x^4-3x^3}{5x^4+7x^2} &=\lim_{x\to0^+}\dfrac{\left(\rule{0pt}{10pt}6x^4-3x^3\right)}{\left(\rule{0pt}{10pt}5x^4+7x^2\right)} \,\dfrac{\;\dfrac{1}{x^2}\;}{\dfrac{1}{x^2}} \\ &=\lim_{x\to0^+}\dfrac{6x^2-3x}{5x^2+7} =0 \end{aligned}\] In the last step, the denominator is still non-zero, \(7\), but now all the terms in the numerator goes to \(0\).

In the last two examples, if we had divided by the largest term in the numerator, then the denominator would not have become a constant and we could not have applied the ordinary Quotient Law.