Question Video: Finding the Limit of a Function Involving Absolute Value Mathematics

Video Transcript

If the function 𝑓 of 𝑥 is equal to the absolute value of 𝑥 plus one minus the absolute value of 𝑥 minus five, find the limit as 𝑥 approaches negative one of 𝑓 of 𝑥.

We’re given a function 𝑓 of 𝑥 and asked to evaluate the limit as 𝑥 approaches negative one of 𝑓 of 𝑥. And the first thing we should always check is “Can we evaluate this limit by using direct substitution?” And to determine this, we need to take a look at our function 𝑓 of 𝑥. It’s the absolute value of 𝑥 plus one minus the absolute value of 𝑥 minus five. And at first, we might be worried whether we can use direct substitution to evaluate this limit. But we can actually rewrite this into a form which we will be able to see that we can use direct substitution.

The first thing we need to remember is the absolute value of a number will tell us the magnitude of that number. In other words, we don’t mind whether our input 𝑦 is positive or negative. We just want to know the size of 𝑦. But there’s actually another way we could write this. Consider 𝑦 squared. If 𝑦 is negative, then 𝑦 squared will be positive. And if 𝑦 is positive, 𝑦 squared will also be positive. So, squaring 𝑦 removes its sign. If we then take the square root of this, we can see that these are the same. The absolute value of 𝑦 is the positive square root of 𝑦 squared. And we will rewrite this as 𝑦 squared raised to the power of one-half.

So, the absolute value of 𝑦 is actually just the composition of two power functions. And we know we’re allowed to evaluate the limits of power functions by using direct substitution. So because the absolute value is the composition of power functions, this means we can also evaluate the limit of absolute values by using direct substitution. We’re now almost ready to evaluate our limit by using direct substitution. There’s only two small points we need to make.

First, we’re not taking the absolute value of 𝑥. In both of our absolute values, we’re taking the absolute value of a linear function. But remember, we can evaluate the limit of any polynomial by using direct substitution. So, each of these two terms is the composition of two functions which we can evaluate by using direct substitution. Finally, we’re taking the difference of these two functions. But we can also evaluate this by direct substitution.

So, we’ve shown the function inside of our limit is the sum, difference, and composition of functions which we can evaluate by using direct substitution. This means we can just substitute 𝑥 is equal to negative one into our limit. This gives us the absolute value of negative one plus one minus the absolute value of negative one minus five. Simplifying the expressions inside of our absolute values symbol, we get the absolute value of zero minus the absolute value of negative six. Of course, the absolute value of zero is zero, and the absolute value of negative six is equal to six. So, this simplifies to give us negative six.

Therefore, by proving we could evaluate this limit by using direct substitution, we were able to show the limit as 𝑥 approaches negative one of the absolute value of 𝑥 plus one minus the absolute value of 𝑥 minus five is equal to negative six.