A *rational function* is the ratio of two functions, f(x) and
g(x).

To find lim_{x−>c} f(x)/g(x), where f(x) and g(x) are continuous
functions at x=c, then first look at

f(c)/g(c). If g(c) is not zero, then the limit is simply f(c)/g(c).

If g(c) is zero, then if f(c) is not zero, the limit is undefined

Finally, if both f(c) and g(c) are zero, then the rational function has the *indeterminate
form* of 0/0, and L'Hopital's Rule can be used to find the limit of the
rational function.

### L'Hopital's Rule

If f(x) and g(x) are differentiable on the interval (a,b) which contains c,
except possibly at c itself, and lim_{x−>c}f(c) = lim_{x−>c}g(c) = 0, then:

lim_{x−>c} (f(x)/g(x)) = lim_{x−>c} (f'(x)/g'(x)),

as long as f'(x) and g'(x) don't change sign infinitely often in a
neighborhood of c.

From this version of the rule, it's possible to prove other variants of it,
for one-sided limits, limits as x approaches plus or minus infinity, or when the
limits of f and g are both infinite. In all cases,

lim (f(x)/g(x)) = lim (f'(x)/g'(x))

See the Mathworld article on L'Hospital's
Rule (an alternative spelling) for more information.

### Example

Evaluate lim_{x−>0} sin(2x)/5x

Since the lim_{x−>0} sin(2x)/5x gives the
indeterminate form, 0/0, you have to take the derivative of the numerator,
sin(2x), and the denominator, 5x, and then divide:

d/dx sin(2x) = 2cos(2x)

d/dx 5x = 5

Set up the quotient:

2cos(2x)/5

As x−>0, this approaches 2/5.

### Internet references

Karl's Calculus Tutor: L'Hopital's
Rule (and the Cheshire Cat's Grin)

Analyze Math: Limits,
Indeterminate Form and L'Hopital's Rule

Mathworld: L'Hospital's
Rule

### Related pages in this website

Limits

Definition of Continuous

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Graeme McRae.