We want to find the limit
as x approaches zero of sine eight x
divided by tangent three x,
but as x approaches zero, the numerator approaches sine zero
and the denominator approaches tangent zero,
which are both equal to zero.
So we say this limit has the indeterminant form
of zero divided by zero,
but because of this indeterminant form,
we can apply L'Hopital's rule to help us
determine this limit.
Where if we have a limit that fits one of these
indeterminant forms, then the limit of this quotient
is equal to the limit of the quotient
of the derivatives of f and g.
Which means this limit is equal to the limit
as x approaches zero of the derivative of sine eight x,
divided by the derivative of tangent three x.
The derivative of sine eight x would be equal to
cosine eight x times eight
or eight cosine eight x.
The derivative of tangent three x would be equal to
secant squared three x times three
or three secant squared three x.
In this form notice how as x approaches zero,
the numerator approaches eight times cosine zero
and the denominator approaches
three times secant squared zero
and because cosine zero and secant zero
are both equal to one, this is equal to eight times one,
divided by three times one squared or just 8/3.
Let's verify this graphically.
Here's where x equals zero.
As we approach zero from the right
and zero from the left, notice how we're approaching
the same function value which we can see would be 8/3
verifying our limit.
Now for our next example, notice how we don't
have a quotient and therefore, in it's current form
this limit is not going to be one of the indeterminant forms
given here where we can apply L'Hopital's rule.
So for the first step, we need to be able to recognize
that secant theta is equal to one over cosine theta
and therefore, we can rewrite this as
the limit as x approaches pi over two
of negative two cosine seven x,
divided by cosine five x.
Again, secant five x is equal to one over cosine five x.
And notice in this form, as x approaches pi over two,
cosine seven x approaches cosine seven pi over two,
which approaches zero and cosine five x approaches
cosine five pi over two, which also approaches zero.
So now we do have the indeterminant form
of zero divided by zero, so we can apply L'Hopital's rule.
So this limit is equal to
the limit as x approaches pi over two,
of the derivative of negative two cosine seven x,
which would be negative two
times the derivative of cosine seven x, not listed here,
would be equal to negative sine seven x times seven
or negative seven sine seven x.
The derivative of cosine five x would be equal to
negative sine five x times five
or negative five sine five x.
So as x approaches pi over two,
notice how the numerator is going to approach
positive 14 times sine of seven pi over two
and the denominator is going to approach
negative five times sine of five pi over two.
Now let's go to the unit circle to determine
these two sine function values.
On the unit circle, sine theta is equal to y.
So for sine five pi over two radians,
this would be the initial side.
We'll rotate counterclockwise five pi over two radians,
where one revolution would be two pi radians,
which is equal to four pi over two radians.
So if we rotate pi over two radians more, here,
we'd be at five pi over two radians.
And since the y-coordinate is equal to one,
we know that sine five pi over two radians is equal to one
and then for sine seven pi over two radians,
if this is five pi over two radians,
we need to rotate pi radians more,
which would take us to here.
So sine seven pi over two radians is equal to negative one.
So going back to our limit,
we would have 14 times negative one,
divided by negative five,
times positive one,
which is equal to negative 14 divided by negative five
or positive 14/5, which would be our limit.
Now let's verify this graphically.
Here's where x equals pi over two radians,
as we approach from the right and we approach from the left
we are approaching the same function value, here,
which would be 14/5, which is our limit.
I hope you found this helpful.