Video Transcript
Determine the limit as 𝑥
approaches two of the natural logarithm of 𝑥 minus one all divided by 𝑥 minus
two.
We’re asked to evaluate the limit
of the quotient of two functions, and the first thing we should check is if we can
evaluate this limit by direct substitution. And in this case, our limit is as
𝑥 is approaching two. We know our numerator is continuous
when 𝑥 is equal to two and our denominator is also continuous when 𝑥 is equal to
two. So, we can attempt to evaluate this
limit by direct substitution.
So, we’ll substitute 𝑥 is equal to
two into the function inside of our limit, giving us the natural logarithm of two
minus one all divided by two minus two. However, the natural logarithm of
two minus one simplifies to give us the natural logarithm of one, which we know is
equal to zero. And two minus two is also equal to
zero. So, this simplified to give us the
indeterminant form of zero divided by zero. This means we’re going to need to
try a different method to evaluate this limit. And because we got an indeterminant
form by direct substitution, and we know how to differentiate both the numerator and
denominator of the function inside of our limit, we can try using L’Hôpital’s
rule.
L’Hôpital’s rule tells us if 𝑓 and
𝑔 are differentiable functions around 𝑎 and 𝑔 prime of 𝑥 is not equal to zero
around 𝑎, except possibly when 𝑥 is equal to 𝑎, and both the limit as 𝑥
approaches 𝑎 of 𝑓 of 𝑥 and the limit as 𝑥 approaches 𝑎 of 𝑔 of 𝑥 are equal to
zero, then the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to the
limit as 𝑥 approaches 𝑎 of 𝑓 prime of 𝑥 divided by 𝑔 prime of 𝑥.
In other words, under these
conditions, we can turn a question about the limit of the quotient of two functions
into a question about the limit of the quotient of their derivatives. And sometimes, the derivatives can
be simpler to work with. And it’s also worth pointing out
this is only true if this limit exists or is equal to positive or negative ∞. So, we need to check we can apply
L’Hôpital’s rule to the limit given to us in the question.
First, we see that our limit is as
𝑥 is approaching two. So, we’ll set our value of 𝑎 equal
to two. And in fact, we can update
L’Hôpital’s rule with 𝑎 set to be two. Next, we’ll set 𝑓 of 𝑥 to be the
function in our numerator and 𝑔 of 𝑥 to be the function in our denominator. So that’s 𝑓 of 𝑥 is the natural
logarithm of 𝑥 minus one and 𝑔 of 𝑥 is 𝑥 minus two. We can now start checking our
prerequisites for L’Hôpital’s rule.
First, we need to check that both
𝑓 and 𝑔 are differentiable around 𝑥 is equal to two. And we can see this directly from
the definitions of 𝑓 of 𝑥 and 𝑔 of 𝑥. We can see that 𝑓 of 𝑥 is the
natural logarithm of 𝑥 minus one. We can in fact find the derivative
of this expression. And we can see it’s differentiable
around 𝑥 is equal to two. And the same is true for 𝑔 of
𝑥. It’s a linear function, so it’s
differentiable for all real values of 𝑥. In particular, it will be
differentiable around 𝑥 is equal to two.
Our next prerequisite says that 𝑔
prime of 𝑥 is not allowed to be equal to zero around 𝑥 is equal to two, although
𝑔 prime of two is allowed to be equal to zero. The easiest way to do this is to
find an expression for 𝑔 prime of 𝑥. That’s the derivative of the linear
function 𝑥 minus two. We know this will just be the
coefficient of 𝑥, which is one. So, 𝑔 prime of 𝑥 is the constant
value of one. This isn’t equal to zero for any
value of 𝑥. So, our second prerequisite is also
true.
Now, all we need to check is the
limit as 𝑥 approaches two of 𝑓 of 𝑥 and the limit as 𝑥 approaches two of 𝑔 of
𝑥 is equal to zero. And we could do this directly by
using direct substitution; however, we’ve already evaluated both of these
limits. When we initially tried direct
substitution on our limit, consider what we did in just the numerator. We substituted 𝑥 is equal to two
into our numerator, giving us the natural logarithm of two minus one, which we
showed was equal to zero. So, the working out in our
numerator tells us the limit as 𝑥 approaches two of 𝑓 of 𝑥 is equal to zero by
using direct substitution.
The same is true if we just look at
the denominator of this working. We substitute 𝑥 is equal to two
into our denominator, showing the limit as 𝑥 approaches two if 𝑔 of 𝑥 is equal to
zero. And of course, we knew both 𝑓 and
𝑔 were continuous at 𝑥 is equal to two. So, we can evaluate the limits by
direct substitution. This means instead of evaluating
the limits in our question, we can instead try evaluating the limit as 𝑥 approaches
two of 𝑓 prime of 𝑥 divided by 𝑔 prime of 𝑥.
However, to do this, we’re first
going to need to find an expression for 𝑓 prime of 𝑥. 𝑓 prime of 𝑥 will be the
derivative of the natural logarithm of 𝑥 minus one with respect to 𝑥. We could evaluate this by using the
chain rule. However, we could also recall for
any real constant 𝑎, the derivative of the natural logarithm of 𝑥 plus 𝑎 with
respect to 𝑥 is equal to one divided by 𝑥 plus 𝑎. In our case, the value of 𝑎 is
negative one. So, we get that 𝑓 prime of 𝑥 is
equal to one over 𝑥 minus one.
We’re now ready to apply
L’Hôpital’s rule to the limit given to us in the question. It’s equal to the limit as 𝑥
approaches two of 𝑓 prime of 𝑥 divided by 𝑔 prime of 𝑥 provided this limit
exists or is equal to positive or negative ∞. The next thing we need to do is
substituting our expressions for 𝑓 prime of 𝑥 and 𝑔 prime of 𝑥. This gives us the limit as 𝑥
approaches two of one over 𝑥 minus one all divided by one. And of course, dividing by one
isn’t going to change our expression. So, this simplifies to give us the
limit as 𝑥 approaches two of one over 𝑥 minus one.
This is the limit of a rational
function. So, we can attempt to evaluate this
by direct substitution. Substituting 𝑥 is equal to two
into our rational function gives us one over two minus one, which we can evaluate is
equal to one. And this is our final answer.
Therefore, by using L’Hôpital’s
rule, we were able to show the limit as 𝑥 approaches two of the natural logarithm
of 𝑥 minus one all divided by 𝑥 minus two is equal to one.
