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 |

.

**.**

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

We find the limit as

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:

.

.

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, |

.

.

**Other Indeterminate Forms 0 ^{ 0}, or (exponential forms)**

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*.

.

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 (*x ^{ 2}*) and then set

.

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

. |
||

.

1) | |

- so,
y = e^{º} = º | |

. | |

. | |

We apply l'Hopital's rule to
we get |

( Calculus II MathRoom Index )

(*all content **© MathRoom Learning Service; 2004 - *).