7. Computing Limits
d. Limits at Infinity
3. When Limit Laws Don't Apply
Limits without Laws - Limit Tricks
a. 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 or even more general functions.
A generalized polynomial is a sum of terms each of which is a coefficient times the variable to a numerical power, not necessarily non-negative integers. If all the powers are non-negative integers, then it is a polynomial.
Compute \(\displaystyle \lim_{x\to\infty}\dfrac{6x^4-3x^2}{5x^4+7x^2}\).
If we plug \(x=\infty\) into the numerator and denominator, we see this has the indeterminate form \(\dfrac{\infty}{\infty}\). To simplify this, we first need to identify the largest term in the denominator. We don't need to look at the coefficients \(5\) and \(7\). For large \(x\), \(x^4\) is bigger than \(x^2\). So we divide numerator and denominator by \(x^4\): \[\begin{aligned} \lim_{x\to\infty}\dfrac{6x^4-3x^2}{5x^4+7x^2} &=\lim_{x\to\infty}\dfrac{\rule[-6pt]{0pt}{12pt}(6x^4-3x^2)}{\rule{0pt}{11pt}(5x^4+7x^2)} \,\dfrac{\;\dfrac{1}{x^4}\;}{\dfrac{1}{x^4}} \\ &=\lim_{x\to\infty}\dfrac{6-\dfrac{3}{x^2}}{5+\dfrac{7}{x^2}} =\dfrac{6}{5} \end{aligned}\]
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 \(6x^4\) became \(6\) and the \(5x^4\) became \(5\) which are constants. The \(3x^2\) became \(\dfrac{3}{x^2}\) and the \(7x^2\) became \(\dfrac{7}{x^2}\) which go to \(0\) as \(x\) goes to \(\infty\).
Compute \(\displaystyle \lim_{x\to\infty}\dfrac{\;\dfrac{2}{x^2}+\dfrac{3}{x^3}\;}{\dfrac{4}{x^2}+\dfrac{2}{x^4}}\).
For \(x\) close to \(\infty\), which is larger: \(\dfrac{1}{x^2}\) or \(\dfrac{1}{x^4}\)?
\(\displaystyle \lim_{x\to\infty}\dfrac{\;\dfrac{2}{x^2}+\dfrac{3}{x^3}\;}{\dfrac{4}{x^2}+\dfrac{2}{x^4}} =\dfrac{1}{2}\)
If we plug \(x=\infty\) into the numerator and denominator, we see this has the indeterminate form \(\dfrac{0}{0}\). To simplify this, we need to identify the largest term in the denominator. If \(x\) is large, \(\dfrac{1}{x^4}\) has more in the denominator than \(\dfrac{1}{x^2}\). So \(\dfrac{1}{x^2}\) is larger. So we divide numerator and denominator by \(\dfrac{1}{x^2}\), which is the same as multiplying by \(x^2\): \[\begin{aligned} \lim_{x\to\infty}\dfrac{\;\dfrac{2}{x^2}+\dfrac{3}{x^3}\;}{\dfrac{4}{x^2}+\dfrac{2}{x^4}} &=\lim_{x\to0}\dfrac{\;\dfrac{2}{x^2}+\dfrac{3}{x^3}\;}{\dfrac{4}{x^2}+\dfrac{2}{x^4}} \,\dfrac{\rule[-6pt]{0pt}{10pt}x^2}{\rule{0pt}{12pt}x^2} \\ &=\lim_{x\to0}\dfrac{2+\dfrac{3}{x}}{\;4+\dfrac{2}{x^2}\;} =\dfrac{1}{2} \end{aligned}\]
If we had multiplied the numerator and denominator by \(x^4\), we would have gotten \(\displaystyle \lim_{x\to0}\dfrac{2x^2+3x}{4x^2+2}\) which would now have been \(\dfrac{\infty}{\infty}\). On the other hand, if we had multiplied the numerator and denominator by \(x^3\), we would have gotten \(\displaystyle \lim_{x\to0}\dfrac{2x+3}{4x+\dfrac{2}{x}}\) which would also have given \(\dfrac{\infty}{\infty}\). Neither gives an obviously finite number.