7. Computing Limits

d. Limits at Infinity

3. When Limit Laws Don't Apply

Limits without Laws - Limit Tricks

b. 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 \(\displaystyle \lim_{x\to\infty}\left(x^2-\dfrac{x^4}{x^2+1}\right)\).

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