Welcome to a video on L'Hopital's Rule, part 1.
The goal this video is to determine limits of indeterminant form using L'Hopitals Rules.
L' Hopital's Rule, also called Bernoulli's Rule,
uses derivatives to evaluate limits involving indeterminant forms.
These are the forms we're talking about,
when we refer to indeterminant forms.
So, you can see in most cases will be dealing with rational functions.
Now, we have encountered limits involving indeterminant form before.
For example, if we have the limit as X approaches zero,
of X squared minus three X,
divided by X square plus two X,
we were able to buy with this limit using
algebraic techniques.
Both the numerator and denominator can be factored.
So, we would have the limit
as X approaches zero of
X,
times X minus three,
divided by X,
times X plus two.
And now, we can see there's a common factor of X here,
which would simplify out,
remember; Graphically, these common factors
create holes in the function,
Remember, holes do not affect limits.
So, now as X approaches zero. We can perform direct substitution.
We would have zero minus three over zero plus
two or negative three halves.
Just to verify this,
if you look at it graphically,
we're reproaching equal zero. Which is this
vertical line here
and we can visually see as we approach from the right
and from the left,
we are approaching a Y value of negative three halves.
And L' Hopital's Rule gives us
a different technique for evaluating limits of indeterminant form
the L'Hopital's rule states the limit of a quotient
is equal to the limit
of the quotient of the derivatives,
under special conditions.
And, here the special conditions
f(x) and g(x) are differentiable are open interval
I
containing C
except possibly at C.
Notice, that C is a value, that were approaching.
We also assume the derivatives of g(x) is not equal to zero.
for all X in the interval
again except possibly at C.
So, if these conditions are met.
And, the limit of the quotient as X approaches C
produces indeterminant form,
then the limit this quotient
is equal to the limit
of the quotient of the derivatives.
and this is also true for a one-sided limits
and this rule can be applied more than once.
So, let's go ahead and give it a try.
Now the limits we look at do meet the conditions
specified by L'Hopitals Rule
but we still should check to make sure it's in the indeterminant form
as stated before.
So, here we have the same limit that we just looked at,
using algebraic technique
but as X approaches zero this
is in the indeterminate form zero over zero.
So, we can apply L'Hopital's Rule.
which means the summit would be equal to the limit
of the quotient of the derivatives.
So, the derivative of X squared minus three X would be two X minus three
the derivative of X squared plus two X should be two X plus two.
Now in this form, we can perform direct substitution
as X approaches zero
we have zero minus three
in the numerator or negative three
In denuminator it would be zero plus two
which is equal to two,
so we get the same limit of negative three halves,
this time using L'Hopital's Rules.
For this next limit,
our numerator and denominator are also approaching zero
Lets apply L'Hopital's Rule
the derivative of numerator would be two,
minus the derivative of sine two X.
that would be cosine two X times two.
applying the chain rule
or just to cosine two X
the derivative of the denominator would be one.
Plus,
two cosine two X.
Now, as X approaches zero,
cosine two X approaches one
so we have two times, two times, one
allover
one plus
two times again cosine two X approaches one
as X approaches zero.
Looks like we have zero over three, which is equal to zero.
let's look at a few more examples here,
Limit as X approaches infinity of natural log X divided by X.
This should be in the form of infinity over infinity,
which is indeterminant
So lets apply L'Hopital's Rule
the derivative of natural log X would be one over X
Of course derivative of X would be one
So, the limit as X approaches infinity of one over X
our numerator is fixed,
Now, denominator is increasing
and when this is the case
this limit will equal zero.
this next limit isn't indeterminate form of zero over zero
as X approaches zero.
So, again lets apply L'Hopital's Rule,
the derivative of sine five X would be cosine five X times five.
Or five cosine five X
what's the derivative of threeX; would be three
and as X approaches zero cosine five X approaches one
So, we have five times one, or five
over three
so this limit is equal to five-thirds.
Let's go take a look at one more example for this video
Then, we'll look at some different types in part two.
The current former this limit does not fit the indeterminate form discussed
before
as X approaches pi over four,
secant two X,
approaches infinity.
And, one minus tangent X
approaches zero,
as X approaches pi over four.
So, sometimes we have to manipulate the function
in order to apply the L'Hopital's Rule.
What I mean by that, is we can rewrite
secant two X
as one over cosine two X
to this will be the same as X approaches pi over four
of one mine is tangent X
divided by
cosine two X
now both the numerator and denominator approach zero.
as X approaches pi over four
so we can apply L'Hopital's rules rule
the derivative of numerator would be zero minus
secant square X or negative secant square X
the derivative of cosine two X would be negative sine two X times two
or negative two sine two X,
Again, we applied the Chain Rule, here.
See what happens as X approaches pi over four.
let's go and start the denominator
as X approaches pi over four.
Two X approaches pi over two,
sine of pi over two is equal to one.
So, we'd have negative two times one in the nominator
And, lets take a look at a reference triangle for the numerator.
Pi over four is a forty-five degree reference triangle.
Three side would be one, one,square two.
Remember secant X is a reciprocal cosine X
So, cosine pi over four would be one over square root two
which means secant pi over four
will be equal to square-root two over one
or just square-root two.
So, we've negative and then square root two squared
So, we'd have negative two over negative two.
So, this limit is equal to one.
This is gonna do it, for part one.
In part two, we'll take a look at more types have limits like this
Where, we have to perform some extra steps before we can apply L'Hopital's Rule
But, before we go let's go ahead and summarize
these are all the indeterminate forms of limits.
And, these here are considered determinant forms
So, you may wanna pause and take a look at these
but we will review these again in part two.
I hope you found this helpful and thank you for watching