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

# Question Video: Finding the Limit of a Function Involving Absolute Value Mathematics • Second Year of Secondary School

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If the function π(π₯) = |π₯ + 9| β |π₯ β 12|, find lim_(π₯ βΆ 12) π(π₯).

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### Video Transcript

If the function π of π₯ is equal to the absolute value of π₯ plus nine minus the absolute value of π₯ minus 12, find the limit as π₯ approaches 12 of π of π₯.

In this question, weβre given a function π of π₯ and asked to evaluate the limit as π₯ approaches 12 of π of π₯. And the first thing we should always do when weβre asked to evaluate a limit like this is ask the question, can we do this by using direct substitution? To do this, letβs start by looking at our function π of π₯. The first thing we notice is the expressions inside of our absolute value symbols are polynomials. And we know polynomials are allowed for direct substitution. But we can see weβre taking the absolute value of these polynomials. So we need to determine if we can use direct substitution for absolute values.

And in fact, thereβs a trick to see this. Consider the square root of π₯ plus nine all squared. By squaring π₯ plus nine, we remove its sign. And then, by taking the square root, we take it back to its original value. In other words, this is exactly equal to the absolute value of π₯ plus nine. But by using our laws of exponents, instead of taking the square root of this entire expression, we can instead raise it to the power of one-half. So we can rewrite the absolute value of π₯ plus nine as the composition of power functions. And we know that power functions are allowed for direct substitution. And we also know the composition of two functions which are allowed for direct substitution is also allowed for direct substitution.

And of course, we can do the exact same thing for the absolute value of π₯ minus 12. We can write this as π₯ minus 12 all squared all raised to the power of one-half. But the key thing to take away here is we can evaluate the limit of the absolute value of π₯ plus nine and the limit of the absolute value of π₯ minus 12 by using direct substitution. So we donβt need our more complicated definition for π of π₯; we can remove this. This means weβve proven that π of π₯ is the difference between two functions which we can use direct substitution. And therefore, this means we can evaluate the limit as π₯ approaches 12 of π of π₯ by using direct substitution.

So now weβre trying to evaluate our limit by using direct substitution. We just substitute π₯ is equal to 12 into our expression for π of π₯. This gives us the absolute value of 12 plus nine minus the absolute value of 12 minus 12. Simplifying the expressions inside of our absolute value symbol, we get the absolute value of 21 minus the absolute value of zero. And of course, 21 is positive, so its absolute value is equal to itself. And the absolute value of zero is just equal to zero, so this simplifies to give us 21.

Therefore, by showing that we could evaluate the limit of π of π₯ by using direct substitution, we were able to show the limit as π₯ approaches 12 of the absolute value of π₯ plus nine minus the absolute value of π₯ minus 12 is equal to 21.

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