### Video Transcript

Given π of π₯ is equal to the
modulus of π₯ plus 11 minus the modulus of π₯ minus 18, find the limit as π₯ goes to
four of π of π₯.

Now our list of conditions under
which direct substitution works does not include the modulus function. However, we can think of the
modulus function in another way. We can write the modulus of π₯ as
the square root of π₯ squared. Since finding the modulus of a
number is simply taking the absolute value of that number and taking the square and
then the root of a number will also give us the absolute value of that number. The square root of π₯ squared can
also be written as π₯ squared to the power of one-half.

π₯ squared and π₯ to the power of a
half are both power functions. We know that we can apply direct
substitution to power functions. And π₯ squared to the power of a
half is simply a composition of two power functions. Therefore, we can also apply direct
substitution to this. From this, we can see that we can
apply direct substitution to the function mod π₯.

Now weβre ready to consider the
function π of π₯. We will consider the modulus of π₯
plus 11 and the modulus of π₯ minus 18. Here weβre simply taking the
moduluses of two polynomial functions, which are π₯ plus 11 and π₯ minus 18. This is simply a composition of a
polynomial function and a modulus function. And since we know we can apply
direct substitution to both polynomial functions and modulus functions, we know that
we can also apply direct substitution to these composite functions.

Now π of π₯ is simply the
difference of these two composite functions. Therefore, we can also apply direct
substitution to π of π₯. So the limit as π₯ approaches four
of π of π₯ is equal to π of four, which is also equal to the modulus of four plus
11 minus the modulus of four minus 18, or mod 15 minus mod of negative 14, which is
also equal to 15 minus 14. From this, we find that the
solution to this question is simply one.