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

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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 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

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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.

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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!

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

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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,

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These are really wierd indeterminate forms. Let's look at 0 0.
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 0 0, 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.

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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.

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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.

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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 (x 2) and then set x = .

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

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 1) so, y = eº = º . . We apply l'Hopital's rule to ln y we get ln y = 1 therefore y = e.

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