We are asked in which limits below
can we use L'Hopital's Rule?
L'Hopital's Rule, also called Bernoulli's Rule,
uses derivatives to evaluate limits
involving indeterminant forms.
The most common indeterminant form is zero divided by zero,
however if a limit is in any of the indeterminant forms
listed here, L'Hopital's Rule does apply
which states the limit as x approaches c of f of x
divided by g of x equals the limit as x approaches c
of f prime of x divided by g prime of x.
And the idea is that this limit will be easier
to evaluate than the original limit.
But again, we can only apply L'Hopital's Rule
if the limit is in one of these indeterminant forms.
If it's not in one of these forms,
L'Hopital's Rule does not apply.
Going back to our exercise, we are only checking
to see if L'Hopital's Rule applies.
We will not be determining the limits.
Looking at the first limit, we have the limit
as x approaches zero of sin two x divided by two x.
As x approaches zero, sin two x approaches sin zero.
And as x approaches zero, two x approaches zero.
And because sin zero equals zero, this limit is in
the form of zero divided by zero, which is one of
the indeterminant forms in which L'Hopital's Rule applies
and therefore we could apply L'Hopital's Rule
to help determine this limit.
Now if needed, we can check that sin zero equals zero
by using the unit circle or by looking at the graph
of f of x equals sin x.
On the unit circle, sin data is equal to the y coordinate,
and here's zero radians on the unit circle.
Notice how at this point on the unit circle,
the y coordinate is zero verifying sin zero equals zero.
Or by looking at the graph of the basic sin function,
which is graphed here, notice that at zero radians,
the function value is zero.
Next we have the limit as x approaches pi over six
of sin x divided by the quantity of pi over six minus x.
As x approaches pi over six, sin x approaches
sin pi over six.
And the quantity pi over six minus x approaches zero
as x approaches pi over six.
And sin pi over six is equal to 1/2,
and therefore this limit is in the form
of 1/2 divided by zero, which does not fit
in one of the indeterminant forms for L'Hopital's Rule,
and therefore L'Hopital's Rule does not apply.
And again, if we needed to, we can verify sin pi
over six equals 1/2 by going back to the unit circle.
Here's pi over six.
At this point on the unit circle, notice how the y
coordinate equals one half, which means sin pi
over six is one half.
Next we have the limit as x approaches zero
of the quantity one minus e to the x divided by sin five x.
As x approaches zero, the numerator approaches one minus e
to the zero.
And the denominator approaches sin zero.
E to the zero was equal to one, and therefore
the numerator approaches zero, and we already know
that sin zero is equal to zero.
This limit is in the indeterminant form of zero divided
by zero, and therefore L'Hopital's Rule does apply.
And then finally we have the limit as x approaches pi over
two of cosine x divided by the quantity two x minus pi.
As x approaches pi over two, cosine x
approaches cosine pi over two.
In the denominator, as x approaches pi over two,
two x minus pi approaches two times pi over two
minus pi.
Cosine pi over two is equal to zero.
Again we can verify this on the unit circle,
as well as looking at the graph
of the basic cosine function.
On the unit circle, cosine data is equal to x.
Here's pi over two.
At this point on the unit circle, the x coordinate is zero
verifying cosine pi over two equals zero.
Or looking at the graph of the basic cosine function,
which we should recognize here, notice cosine pi
over two equals zero.
In the denominator, two times pi over two is equal to pi.
Pi minus pi is equal to zero.
This limit is also in the indeterminant form of zero
divided by zero, and therefore L'Hopital's Rule does apply.
I hope you found this helpful.