When faced with a limit which becomes 0/0 (undefined), we are faced with several choices. Sometimes, we can factor out the term which is vanishing (such as lim_x->0 2x/3x) but many times, such as in lim_x->0 sin(x)/x, one cannot explicitly cancel the factor x->0. If f(x) and g(x) both vanish at a point x=a, then

lim_x->a f(x)/g(x) = 0/0

is called an indeterminate form of type 0/0.

A related problem is when one gets infinity/infinity as an result, such lim_x->infinity ln(x)/x. This is called an indeterminate form of type infinity/infinity . In both of these cases, we can convert the indeterminate limit into an equivalent limit, which may be defined.

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L'Hopital's Rule If f and g are differentiable, and g'(x) does not vanish on an interval containing x=a (except possibly at a itself), and suppose

lim_x->a f(x) = 0 = lim_x->a g(x)

or that

lim_{|x|->infinity} |f(x)| = infinity = lim_{|x|->infinity} |g(x)|

Then

lim_x->a f(x)/g(x) = lim_x->a f'(x)/g'(x)

If f' and g' both vanish (or become infinite), then this process can be repeated.

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A quick derivation of this rule (for a special case) is as follows (if f(a)=g(a)=0)

lim _x->a f(x)/g(x)

= lim _x->a [f(x)-f(a)]/[g(x)-g(a)]

= lim _x->a [(f(x)-f(a))/(x-a)]/[(g(x)-g(a))/(x-a)]

= lim _x->a [(f(x)-f(a))/(x-a)]/ lim _x->a [(g(x)-g(a))/(x-a)]

= f'(a)/g'(a)

= lim _x->a f'(x)/g'(x)

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Another way of looking at this is to approximate f(x) near f(a)=0 by a straight line f(x) ~ f'(a)(x-a), and similarly with g(x) ~ g'(a)(x-a). Then f(x)/g(x) ~ [f'(a)(x-a)]/[g'(a)(x-a)] =f'(a)/g'(a).

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As a quick example, consider lim_x->0 sin(x)/x = lim _x->0 cos(x)/1 = 1. Remember, L'Hopital's rule does not use the quotient rule! Remember, also, to check the functions to see that the assumptions of L'Hopital are valid!

In some cases, one must rearrange terms somewhat. For example, lim _{x-> 0} x ln(x) = lim _{x-> 0} ln(x)/(1/x). Otherwise a term such 0.infinity occurs, which is indeterminate also...

Indeterminate differences of the type infinity-infinity can sometimes reduced to previous forms by rewriting with a common denominator or rationalizing expressions.

Indeterminate powers such as 0⁰, 1^infinity, or infinity⁰ or 0 ^infinity may be reduced to first two forms by

lim _x->a f(x)^g(x) = lim _x->a e^{g(x) ln(f(x))}

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A classic example to check if you understand the concepts is:

lim_{x->0 ⁺} x^x

Last modified Wed Oct 30 22:09:46 CST 1996