5.12.1 Further Limits

l’Hôpital’s rule is a method involving calculus that allows us to find the value of certain limits
Specifically, it allows us to attempt to evaluate the limit of a quotient $\frac{f (x)}{g (x)}$ for which our usual limit evaluation techniques would return one of the indeterminate forms $\frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$ .

STEP 1: Check that the limit of the quotient results in one of the indeterminate forms given above

I.e., check that $\lim_{x \to a} \frac{f (x)}{g (x)} = \frac{f (a)}{g (a)} = \frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$

STEP 2: Find the derivatives of the numerator and denominator of the quotient
STEP 3: Check whether the limit $\lim_{x \to a} \frac{f' (x)}{g' (x)}$ exists
STEP 4: If that limit does exist, then $\lim_{x \to a} \frac{f (x)}{g (x)} = \lim_{x \to a} \frac{f' (x)}{g' (x)}$
STEP 5: If $\lim_{x \to a} \frac{f' (x)}{g' (x)} = \frac{f' (a)}{g' (a)} = \frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$ then you may repeat the process by considering $\lim_{x \to a} \frac{f'' (x)}{g'' (x)}$ (and possibly higher order derivatives after that)

As long as the limits continue giving indeterminate forms you may continue applying l’Hôpital’s rule
Each time this happens find the next set of derivatives and consider the limit again

Use l’Hôpital’s rule to evaluate each of the following limits:

a) $\lim_{x \to 0} \frac{\sin x}{e^{x} - 1}$ .

5-12-1-ib-aa-hl-lhopitals-rule-a-we-solution

b) $\lim_{x \to 0} \frac{x^{3}}{- 2 x + \sin 2 x}$ .

5-12-1-ib-aa-hl-lhopitals-rule-b-we-solution

Some limits of an indeterminate form can also be evaluated using the Maclaurin series for the numerator and denominator
If an exam question does not specify a method to use, then you are free to use whichever method you prefer

Limits of the form $\lim_{x \to a} \frac{f (x)}{g (x)}$ or $\lim_{x \to \infty} \frac{f (x)}{g (x)}$ may sometimes be evaluated by using Maclaurin series
Usually this will be in a situation where attempting to evaluate the limit in the usual way returns an indeterminate form $\frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$ .
In such a case:

STEP 1: Find the Maclaurin series for $f (x)$ and $g (x)$
STEP 2: Rewrite $\frac{f (x)}{g (x)}$ using the Maclaurin series in the numerator and denominator
STEP 3: Use algebra to simplify your new expression for $\frac{f (x)}{g (x)}$ as far as possible
STEP 4: Evaluate the limit using your simplified form of the expression

Use Maclaurin series to evaluate the limit

$\lim_{x \to 0} \frac{x^{3}}{- 2 x + \sin 2}$

5-12-1-ib-aa-hl-maclaurin-series-limits-we-solution

Some limits of an indeterminate form can also be evaluated using l’Hôpital’s Rule
If an exam question does not specify a method to use, then you are free to use whichever method you prefer

IB DP Maths: AA HL

5.12.1 Further Limits