Video Transcript
Discuss the existence of the limit
as 𝑥 approaches two of 𝑓 of 𝑥 given that 𝑓 of 𝑥 is equal to 20 minus two 𝑥
plus 𝑥 times the absolute values of 𝑥 when 𝑥 is greater than negative six and
less than two and 𝑓 of 𝑥 is equal to five times 𝑥 minus two over the square root
of 𝑥 plus two minus two when 𝑥 is greater than two and less than 14.
In this question, we’re given a
piecewise-defined function 𝑓 of 𝑥 and asked to determine if the limit of this
function exists as 𝑥 approaches two. We can note that 𝑥 equals two is
not in the domain of the function. However, it does lie on the
endpoints of the function’s subdomains. To determine the existence of this
limit, we first recall that we say that the limit as 𝑥 approaches two of 𝑓 of 𝑥
exists if both the left and right limit of 𝑓 of 𝑥 exist and are equal. It is worth noting that this is
multiple conditions. However, we usually write this in
shorthand as a single equation showing that the left and right limits are equal.
Therefore, to determine the
existence of this limit, we need to check the left and right limit as 𝑥 approaches
two of 𝑓 of 𝑥. Let’s start with the limit as 𝑥
approaches two from the left. This means that our values of 𝑥
will be less than two. We can see that when 𝑥 is less
than two, 𝑓 of 𝑥 is equivalent to its first subfunction, where it is worth noting
that we can assume 𝑥 is greater than negative six since we are taking a limit. So the values of 𝑥 will approach
two. Since 𝑓 of 𝑥 is equivalent to its
first subfunction as 𝑥 approaches two from the left, we can conclude that their
limits as 𝑥 approaches two from the left will also be equal.
We can then note that this
subfunction is the sum and product of functions whose limits can be evaluated by
direct substitution. This means that we can attempt to
evaluate this limit by direct substitution. We substitute 𝑥 equals two into
the first subfunction to obtain 20 minus two times two plus two times the absolute
value of two. We can evaluate this to obtain
20. Hence, we have shown that the limit
as 𝑥 approaches two from the left of 𝑓 of 𝑥 is 20.
We now need to find the limit as 𝑥
approaches two from the right of 𝑓 of 𝑥. Once again, we can say the values
of 𝑥 will be greater than two, since we are taking the limit as 𝑥 approaches two
from the right. We will also take our values less
than 14 so that 𝑓 of 𝑥 is equivalent to its second subfunction. Thus, the limit as 𝑥 approaches
two from the right of 𝑓 of 𝑥 equals the limit as 𝑥 approaches two from the right
of the second subfunction. If we try to evaluate this limit by
direct substitution, then we see that we obtain the indeterminate form zero divided
by zero. This means we need to evaluate this
limit using a different method.
Since the denominator of the
function is a radical expression, we can simplify this limit by multiplying both the
numerator and denominator by the conjugate of the denominator. We can simplify the denominator by
using a difference of two squares. We get the limit as 𝑥 approaches
two from the right of five times 𝑥 minus two multiplied by the square root of 𝑥
plus two plus two all divided by root 𝑥 plus two all squared minus two squared. We can simplify the denominator
further by noting that the square root of 𝑥 plus two all squared is just 𝑥 plus
two. Then we subtract four to obtain 𝑥
minus two.
We can then note that there is a
shared factor of 𝑥 minus two in the numerator and denominator. We can cancel this shared factor by
noting that we’re taking a limit. So we are only interested in what
happens as the values of 𝑥 approach two from the right, not the value of the
function when 𝑥 equals two. This gives us the limit as 𝑥
approaches two from the right of five times the square root of 𝑥 plus two plus
two.
This is a radical function, so we
can evaluate this limit by direct substitution. Substituting 𝑥 equals two into the
function gives us five times the square root of two plus two plus two. We can evaluate this to get 20. We have now shown that both the
limit as 𝑥 approaches two from the left and right exist, and they are both equal to
20. Hence, the limit exists and equals
20.
