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Determine the limit as 𝑥 tends to four of the square root of 𝑥 plus 12 minus four all over 𝑥 minus four.
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The first thing we should try is direct substitution, just plugging in 𝑥 equals four.
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If we do this, we get the square root of four plus 12 minus four all over four minus four.
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Four plus 12 is 16.
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The square root of 16 is four.
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And four minus four is zero.
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So we get the indeterminate form zero over zero.
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In short, direct substitution doesn’t work here.
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What we need to do is to somehow rewrite this fraction in such a way that we can cancel out a common factor of the numerator and denominator.
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And furthermore, where substituting directly into whatever’s left doesn’t give us zero over zero.
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So how do we rewrite our fraction?
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The reason why this fraction is so difficult is because the numerator has this radical in it.
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It’s hard to find a common factor of the numerator and denominator with this radical here.
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So what can we do about it?
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The key to this question is to multiply both the numerator and denominator by the square root of 𝑥 plus 12 plus four.
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In other words, it’s the numerator but with the minus sign changed to a plus sign.
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Of course, multiplying both numerator and denominator by the same thing gives us an equivalent fraction.
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We can then distribute the terms in the numerator.
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We distribute the terms in the numerator and we see that the two cross terms cancel.
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This is not a coincidence.
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We chose to multiply both numerator and denominator by the square root of 𝑥 plus 12 plus four, in order to make this happen.
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Here, we were thinking about the difference of squares identity, 𝑎 minus 𝑏 times 𝑎 plus 𝑏 is equal to 𝑎 squared minus 𝑏 squared.
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As a result, the only term in the numerator involving the square root of 𝑥 plus 12 has it squared.
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And the square root of 𝑥 plus 12 squared is just 𝑥 plus 12.
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So we have succeeded in getting rid of the problematic square root term in the numerator.
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We have, however, gained a radical in the denominator.
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But we’ll see soon why this isn’t a problem.
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We simplify the numerator further to get 𝑥 minus four which cancels with a factor of 𝑥 minus four in the denominator.
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We’re hoping that this common factor of 𝑥 minus four is the reason that when we substituted in 𝑥 equals four to the fraction, we got the indeterminate form zero over zero.
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And that having canceled this common factor upon directly substituting the value four in again, we’ll get the value of the limit and not the indeterminate form.
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Before we substitute, let’s finish off our algebra.
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We find that the square root of 𝑥 plus 12 minus four all over 𝑥 minus four is equal to one over the square root of 𝑥 plus 12 plus four.
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Okay.
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Let’s clear away our working and get to substituting.
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Here’s what we showed with our algebra.
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And this remains true upon taking limits on both sides.
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We saw before that direct substitution didn’t work on our original fraction but we’re hoping it works on this one.
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Replacing 𝑥 by four, we get one over the square root of four plus 12 plus four.
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Four plus 12 is 16.
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And the square root of 16 is four.
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So we get one over eight.
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So direct substitution did work on the equivalent fraction we found.
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The main trick in this question was to multiply both numerator and denominator by the square root of 𝑥 plus 12 plus four.
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Or equivalently, to multiply by the fraction whose numerator and denominator are both the square root of 𝑥 plus 12 plus four.
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Having done this, we just needed to do some algebra and cancel out the factors of 𝑥 minus four which appeared, before substituting it directly to find the value of the limit.
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I call this a trick because this is not something that anyone would necessarily be expected to come up with themselves.
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This is quite a useful trick.
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And so, you might want to remember that sometimes it’s useful to rationalize the numerator rather than the denominator.