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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 1

Lesson 7: Determining limits using algebraic manipulation

# Limits by rationalizing

Sal finds the limits of (x+1)/(√(x+5)-2) at x=-1 by "rationalizing the denominator" of the expression.

## Want to join the conversation?

• At , Sal introduces the concept to use another function to solve for the limit of g(x). Wouldn't it just be easier to just plug in the limit as x to the simplified version of g(x) without having to use f(x)? •   The point he is making here is perhaps a bit unclear. He's not actually using another function to solve for the limit, per se. He's showing mathematical proofs - that if there are two functions, which are identical except for a discontinuity at a single point, then their limits will be the same, so you can use the limit from the continuous function (which you can get from just plugging in to the equation) for the discontinuous function. So that whole f(x) bit is the mathematical proof that lets you use the technique you just mentioned - just plugging the number into the simplified expression - to get the limit.

I probably would have started with, "If that discontinuity wasn't there, what would the value be? Well, we'd have 'the square root of x plus five, plus two', which at x = -1 would be 4. The discontinuity is there, so the value isn't 4, but we can see that it would be 4 if the hole wasn't there, so the limit is 4." And then given the written out proof that he gives here to show how that is mathematically valid rather than just graphically intuitive.
• Oh! That makes sense! Anything other than 0 divided by 0 is undefined, but 0/0 is indeterminate, which allows you to manipulate it into other forms. I never understood whether or not I'd be justified in manipulating an expression like this into a nicer form. Apparently, it's justified when the numerator is zero. I always see the algebraic manipulations immediately when I see a problem like this, but I always worry it might not be justified...
My question: why IS such manipulation permissible when the numerator is zero? And relatedly, why is this NOT justified when the numerator is non-zero?

Thank you very much for your help! •   This may not be an answer for all cases, but for rational functions (functions with polynomial numerators and denominators) a non-zero number divided by zero indicates the presence of a vertical asymptote in the graph. In most cases, we say the limit of a function does not exist as it approaches an asymptote, so there's no manipulation you could do to find a limit. However (still for rational functions) zero divided by zero indicates the presence of a removable point discontinuity, or "hole." In these cases, though the function does not have a value at that point, it does have a limit, so manipulating it could allow you to find that limit. It is possible this is true of other situations that yield division by zero, but I don't know enough to really say.

This may not be the most valid math (or maybe it is!) but here's another way to think of it: The reason a non-zero number divided by zero is undefined is there is absolutely nothing it could equal under the existing rules and definitions of numbers and math. Turn around an equation such as 2/0 = x and it becomes 0x = 2. There is no number you can multiply by zero and get two! In terms of limits, there is none to be found. But the reason zero divided by zero is undefined is that it could theoretically be any number. Turn around 0/0 = x and it becomes 0x = 0. Anything times zero is zero! In terms of limits, there is a limit there to be found. It's obscured by the 0/0, but some manipulation could reveal it.
• if we just look at " sqrt(x+5) -2 " as the denominator without rationalizing, we know it cannot equal zero. So we solve for zero:
"sqrt(x+5) -2 = 0" --> x = -1

alright, what if we test this? sqrt(-1+5) -2 = 0 --> sqrt(4)-2 =0
BUT, sqrt(4) could equal 2 OR -2.

what's up with this -2? •  As Sal said at , we take principal root, meaning we only take the positive number.
• Why to we want principle square roots only? • For a function to be a function, every input must have a single output. Since each number has two square roots, we need a convention for which one to use. The convention is that the square root sign by itself indicates only the principal (in this case positive) square root. A negative sign in front of the radical would indicate the negative root, and a plus-or-minus sign would indicate both.
• Instead of going through the rationalizing process to find the limit of g(x), is it reasonable to just replace x with something really close to -1 (like -0.99999) so that g(x) doesn't equal to 0/0 and then solving for g(x) using that really close value? • Is there any way for us to say if a function is continuous or not without knowing the limit or graphing it? How does Sal say with such certainty that f(x), for example, is continuous? • We know several things about continuous functions:
1. All polynomials are continuous (including constant and linear functions)
2. The square root function is continuous
3. The sum of continuous functions is continuous
4. The quotient of continuous functions is continuous whenever it's defined
5. The composition of continuous functions is continuous whenever it's defined

We can prove all these things from the epsilon-delta definition of continuity.

Assembling these facts, we can see that f(x) is continuous wherever it's defined (it's undefined, and therefore not continuous, at x= -1).
• But how did sal know that the conjugate would work? Are there situations where it wouldn't work and if so how do I know what these are? • at , instead of multiplying it by sqrt(x+5) +2/sqrt(x+5) +2 , cant we just multiply it by sqrt(x+5) -2/sqrt(x+5) -2 ? • You could do so, but in this case it wouldn't help. Both numerator and denominator would be more complex than before and you would still get 0/0. Furthermore, it wouldn't provide a matched pair of factors in the numerator and denominator, which are necessary for "removing" the discontinuity.

sqrt(x + 5) + 2 is specifically useful because, as the conjugate of sqrt(x + 5) - 2 , it removes the radical in the denominator, and in this case it's the only thing you can multiply by to get an (x + 1) in the denominator, which you can use to "remove" the discontinuity.
• What if the square root expression is in the numerator? • Rationalizing the numerator is also a valid strategy. The purpose of rationalizing is to change the expression into a form that might be easier to work with. From there, you might be able to cancel and simplify until you can use direct substitution. These exercises have removable discontinuities, so any simplification will not change the value of the limit.
• Not sure if this is the video to be looking at, but I've been having a very large issue with this question

Lim (1/sqrt(x))-(1/sqrt(x^2-x))
x->0 