Limits part 2: Functions: Techniques
L'Hôpital's rule
When we studied the calculation rules for limits of functions we came across the following indeterminate forms: \[ \frac{0}{0} \text{ and }\frac{\pm\infty}{\pm\infty} \text.\] An example is \[ \lim_{x\to 0}\frac{\sin(2x)}{\sin(5x)}\text. \] Both the numerator and the denominator converge to \(0\), so we cannot apply the quotient rule for limits of functions. In this paragraph you learn how to apply a theorem to calculate this limit anyway.
L'Hôpitals rule states that under the right assumptions one can differentiate the numerator and the denominator and create with them a new fraction of which the limit is calculated:
L'Hôpital's rule Suppose that \(f\) and \(g\) are function for which \(\lim_{x\to a} \frac{f'(x)}{g'(x)}\) exists. Furthermore assume that \(\displaystyle \lim_{x\to a}f(x)=0\) and \(\displaystyle \lim_{x\to a}g(x)=0\) (both limits equal \(0\)) or that \(\displaystyle \lim_{x\to a}f(x)=\pm\infty\) and \(\displaystyle \lim_{x\to a}g(x)=\pm\infty\) (both limits equal plus or minus infinity).
Then we have: \[\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}\]
L'Hôpital's rule gives the standard limit \[\lim_{x\to 0}\frac{\sin(x)}{x}=\lim_{x\to 0}\frac{\cos(x)}{1}=1\]
Make sure that the conditions of L'Hôpital are satisfied before applying the theorem. L'Hôpital's rule can only be used for the indeterminate forms \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(\frac{\infty}{-\infty}\), \(\frac{-\infty}{\infty}\) and \(\frac{-\infty}{-\infty}\).
We have \[\lim_{x\to 0}\frac{\sin(2x)}{\sin(3x)+1}=0\] because the numerator converges to \(0\) and the denominator converges to \(1\) as \(x\to 0\).
Recklessly applying L'Hôpital yields a different result: \[\lim_{x\to 0}\frac{\displaystyle\frac{\dd}{\dd x}\sin(2x)}{\displaystyle\frac{\dd}{\dd x}(\sin(3x)+1)}=\lim_{x\to 0}\frac{2\cos(2x)}{3\cos(3x)}=\frac{2}{3}\] The conditions weren't satisfied in this case, so we couldn't apply L'Hôpital's rule here.
