# Limit with square root

$$\lim_{x\to16}\frac{4-\sqrt{x}}{16x-x^2}$$

I am not sure what to do, I have tried factoring everything and using both conjugates, neither options gives me anything usable.

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First pull out the obvious factor of $x$ in the denominator to get $$\frac{4-\sqrt x}{x(16-x)}\;.$$ The $x$ in the denominator won’t cause any problems in taking the limit, so focus on the rest: $$\frac{4-\sqrt x}{16-x}\;.$$ Notice that each term in the denominator is the square of the corresponding term in the numerator: $$\frac{4-\sqrt x}{16-x}=\frac{4-\sqrt x}{4^2-(\sqrt{x})^2}\;.$$ That last denominator is the difference of two squares; what do you know about factoring such differences?

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No I forgot the formula. – user138246 Jan 15 '12 at 1:09
@Jordan: David Mitra gave it at the end of his answer. That’s one that you really need to remember: it comes up a lot. – Brian M. Scott Jan 15 '12 at 1:12
For whatever reason I still am not getting the right answer. For the denominator I get $(4+\sqrt{x})(4+\sqrt{x})$ I then cancel out the $(4-\sqrt{x})$ and I am left with $(4+\sqrt{x})$ which is 8. – user138246 Jan 15 '12 at 1:45
@Jordan: That’s fine, but don’t the rest of the expression: I set the $x$ aside to focus on the rest, but it’s still there, so after your cancellation the fraction is $\frac1{x(4+\sqrt{x})}$, and the limit is $\frac1{16\cdot 8}=\frac1{128}$. – Brian M. Scott Jan 15 '12 at 3:41

Let's try to get things to look the same: $${4-\sqrt x\over 16x-x^2}={4(1-{\sqrt x\over4})\over 16x(1-{x\over 16} ) }= { 1-{\sqrt x\over4} \over 4x(1-{x\over 16} ) }.$$ Observe that $({\sqrt x\over 4})^2={x\over 16}$ for $x>0$.

Can you see how to take advantage of the formula $a^2-b^2=(a+b)(a-b)$?

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It’s a perfectly reasonable answer, but I think that introducing the fractions makes it a little uglier than necessary. – Brian M. Scott Jan 15 '12 at 0:44

Alternative Method: (Assuming one knows differentiation)

$$\lim_{x\to16}\dfrac{4-\sqrt{x}}{16x-x^2}=\lim_{x\to16}\frac1x\dfrac{4-\sqrt{x}}{16-x}=\lim_{x\to16}\frac1x\dfrac{\sqrt{16}-\sqrt{x}}{16-x}.$$ By definition, $\lim\limits_{x\to16}\frac{4-\sqrt{x}}{16-x}$ is the derivative of $\sqrt{x}$ evaluated at $x=16$. And since the derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$, it follows that its value is: $\frac{1}{2\sqrt{16}}=\frac{1}{8}$. Thus $\lim\limits_{x\to16}\frac{4-\sqrt{x}}{16-x}=\tfrac18$, hence the value of the limit is $\tfrac1{16}\cdot\tfrac18=\tfrac{1}{128}$.

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Try using L'Hôpital's rule in the case of an indeterminate form.

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It works, but it’s overkill, and Jordan may not yet have reached l’Hospital’s rule in his course. – Brian M. Scott Jan 15 '12 at 0:45

One way is by a rationalizing substitution: \begin{align} u & = \sqrt{x} \\ \\ u^2 & = x \\ \\ \text{As }x\to16, & u \to 4. \end{align} So $$\lim_{x\to16}\frac{4-\sqrt{x}}{16x-x^2} = \lim_{u\to4}\frac{4-u}{16u^2-u^4} = \lim_{u\to 4} \frac{4-u}{u^2(4-u)(4+u)} = \lim_{u\to4} \frac{1}{u^2(4+u)}=\frac{1}{4^2(4+4)}=\frac{1}{128}.$$

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