Limits in indeterminate forms (HL)

This expands upon the discussion in Limits, continuity and differentiability. Here we elaborate on

lim⁡x→a(f(x)+g(x))=L does not imply lim⁡x→af(x)+lim⁡x→ag(x)=L\lim_{x\to a} (f(x) + g(x)) = L \text{ does not imply } \lim_{x\to a} f(x) + \lim_{x\to a} g(x) = L
lim⁡x→a(f(x)⋅g(x))=L does not imply lim⁡x→af(x)⋅lim⁡x→ag(x)=L\lim_{x\to a} (f(x) \cdot g(x)) = L \text{ does not imply } \lim_{x\to a} f(x) \cdot \lim_{x\to a} g(x) = L

Contents

Indeterminate forms

The indeterminate forms are what that would result, if the single limit were split into two. They are

00, ∞∞, 0⋅∞, ∞−∞, 00, 1∞, ∞0\frac00,\, \frac \infty\infty,\, 0 \cdot\infty,\, \infty -\infty,\, 0^0,\, 1^\infty,\, \infty^0

Limits in indeterminate forms may evaluate to any real number, as well as undefined. The l’Hôpital’s rule, possibly through repeated use, can determine the value of such limits.

L’Hôpital’s rule

lim⁡x→af(x)lim⁡x→ag(x) is 00 or ∞∞  ⟹  lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\frac{\displaystyle\lim_{x\to a} f(x)}{\displaystyle\lim_{x\to a} g(x)} \text{ is }\frac00 \text{ or } \frac \infty\infty\implies \lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f^\prime(x)}{g^\prime(x)}

L’Hôpital’s rule is only applicable for limits in one of these two indeterminate forms. It cannot be applied to any arbitrary limit.

L’Hôpital’s rule may be applied multiple times, as long as each intermediate result is indeterminate.

Extension

These are technically not part of the syllabus, but they can be useful to know

For f(x)⋅g(x)f(x) \cdot g(x) in indeterminate form, use

lim⁡x→af(x)⋅g(x)=lim⁡x→af(x)g(x)−1=lim⁡x→ag(x)f(x)−1\lim_{x\to a} f(x)\cdot g(x) = \lim_{x\to a} \frac{f(x)}{g(x)^{-1}} = \lim_{x\to a} \frac{g(x)}{f(x)^{-1}}

Which one goes on the denominator is entirely personal taste, depending on what is easier to differentiate.

For f(x)g(x)f(x)^{g(x)} in indeterminate form

lim⁡x→af(x)g(x)=(e)lim⁡x→ag(x)ln⁡f(x)=(e)lim⁡x→aln⁡f(x)g(x)−1\lim_{x\to a} f(x)^{g(x)} = (e)^{\displaystyle\lim_{x\to a} g(x) \ln f(x)} = (e)^{\displaystyle\lim_{x\to a} \frac{\ln f(x)}{g(x)^{-1}}}

Finally, ∞−∞\infty -\infty is the trickiest but they may have a common factor or it’s possible to convert two fractions into one with a common denominator.

Alternative method

Many limits can be expressed using Maclaurin series, which transforms the limit to a limit of a polynomial. See the bottom of the Maclaurin series page for a practical example.