l'Hopital's Rule

Indeterminate Forms, L'Hopital's Rule

Indeterminate Forms

l'Hopital's Rule

The Indeterminate Form

Other Indeterminate Forms (exponential forms)

The Indeterminate Form

Practice

Solutions

Indeterminate Forms

We can use algebraic techniques to evaluate many types of limits -- even limits at infinity because they limit to either a constant k, or something of the form
all of which can be logically evaluated according to our number systems.

For instance (0/a) = 0 since 0 divided by a constant equals 0.
Similarly, the other fractions in the list can be evaluated at 0 or .

Indeterminate forms such as (0/0) and () pose a problem.
Let's examine the problem with (0/0)
Since numerator = 0, the fraction should = 0
Since denominator = 0, the fraction should =
and, since numerator = denominator, the fraction should = 1
-- which is it?? 0, or 1? -- well, that's the problem
so we use l'Hopital's Rule to evaluate these limits.

l'Hopital's Rule

L'Hopital's rule deals with an exclusive set of indeterminate forms -- the 2 above.
What it says is this:

We want the limit as x approaches c for a quotient of 2 functions.
We substitute x = c and get 0/0 or .
We find the limit as x approaches c for the quotient of the derivatives of the 2 functions.

So, the formal statement of the rule is:

In the lesson on Algebraic limits, we approached this example

algebraically with factor, cancel and substitute.
Using l'Hopital's rule we would get

Sometimes, we must apply the rule repeatedly to get the limit like this

so we apply the rule again.

Here we have 0/0 = ½. Had the denominator been x ³, we
would have to apply the rule 3 times and we'd get 0/6 or 0.

Note: It's very important to verify that we have 0/0 or to start.

Example: We'll do this one without checking the indeterminate form.

is really 2/0 or but we'll pretend we can apply l'Hopital's rule.

We'll get:

The limit is = , not 1, so we must make sure to evaluate f(c)/g(c)
before applying l'Hopital's rule.

If f(c)/g(c) ! 0/0 or l'Hopital's rule doesn't apply!

l'Hopital's rule makes it easy to prove the 2 standard trig limits, namely:

(back to table of contents)

The Indeterminate Form

using l'Hopital's rule.

We turn the product of 2 functions into a quotient because
we know that a $ b = b ÷ (1/a) = a ÷ (1/b)

Note: take advantage of the trig reciprocal functions for things like:

Examples

a) b)

if x = 0, we have 0 $ º, rewrite the limit if x = o/2, we have º $ 0, rewrite the limit

get use l'Hopital's rule get use l'Hopital's rule

so we have so,

(back to table of contents)

Other Indeterminate Forms 0⁰, or (exponential forms)

These are really wierd indeterminate forms. Let's look at 0⁰.
Our number system dictates that the zero power of a constant should = 1
but also that 0 raised to any power should = 0. Again, which is it?

As with all exponential situations, we apply the log function to deal with the exponents.

If y = f(x)^{g (x)} evaluates to 0⁰, or ,

take ln y and use l'Hopital's rule to find .

We must evaluate y and not ln y at the end of the question.

Note: again, take advantage of the trig reciprocal functions
like changing sec x (thing) to thing / cos x because sec x = 1 / cos x.

memory tweak: log x^y = y log x

Example

Now we do a limit that demystifies the irrational constant e.

(back to table of contents)

The Indeterminate Form

When f (c) – g(c) = , and if the functions are fractions, we find
the common denominator, combine the 2 fractions, then apply l'Hopital's rule.

In other cases, we can use the graphs of the functions to evaluate the limit.

For example if we have ,
because we see from the graph of y = ln x that y approaches as x approaches 0.

Thus we know this limit is getting infinitely large.

Note: make sketches of the graphs for the functions in question, to easily evaluate limits.

When there are roots in the expression, we rationalize like this:

We're asked to find which equals .

Since we musn't change values, we multiply by a fraction with a value = 1

For the last step, we use the algorithm for finding horizontal asymptotes -- that is,
we divide through by the highest power of the variable (x²) and then set x = .

(back to table of contents)

Practice

Find the following limits: (Show all your work!)

Solutions

1)

so, y = e^º = º

.

.

We apply l'Hopital's rule to ln y

we get ln y = 1 therefore y = e.

(table of contents)

( Calculus I MathRoom Index )

a)	b)
if x = 0, we have 0 $ º, rewrite the limit	if x = o/2, we have º $ 0, rewrite the limit
get use l'Hopital's rule	get use l'Hopital's rule
so we have	so,

1)


	so, y = e^º = º
.



	.

We apply l'Hopital's rule to ln y we get ln y = 1 therefore y = e.