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I have to find the limit $${\lim_{x \to 49} \frac{\sqrt{x}-7}{x-49} }.$$

I know that I cannot plug in $49$ because that would make the denominator $0$. I was told to rationalize the numerator and I did. This is what I did but I got the incorrect answer: $$\dfrac{\sqrt{x}-7}{x-49}\times\dfrac{\sqrt{x}+7}{\sqrt{x}+7}.$$ I multiplied this out and got $$\dfrac{x-49}{x\sqrt{x}+7x-49\sqrt{x}+343}.$$ Now when I plugged in $49$, the limit came out as $0$ but it was incorrect. Am I missing a step or did I do something wrong? I went over it a couple of times and I cannot catch my mistake.

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@Pambos Definitely, $(-49)(-7)=+343$. But anyway, this is certainly not what one should do at this point. – 1015 Feb 6 '13 at 21:51
@julien Oops, I meant $(x-49)\times(\sqrt x+7)=x\sqrt{x}+7x-49\sqrt{x}\color{red}{-}343$. Of course this not what one should do, I just want to point out that when he plug in $49$, in the (correct) last expression, the answer is $\dfrac00$ and not $0$. – P.. Feb 6 '13 at 21:58
up vote 11 down vote accepted

So close! You actually did too much work. If you leave the denominator as-is you get $$ \dfrac{x-49}{(x-49)(\sqrt{x}+7)} = \dfrac{1}{\sqrt{x}+7} $$

which you can then take the limit of.

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Oh, I guess I just did too much thinking on this problem... I won't be making this mistake again in the future :). Thank you for your answer! – Kot Feb 6 '13 at 21:47

Factorise $x-49$ as $(\sqrt x+7)(\sqrt x-7)$, and divide the numerator and denominator by $\sqrt{x}-7$.

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I don't think this helps to understand the problem in a way for the asker to apply in the future. – Paul Raff Feb 6 '13 at 21:45
Using $a^2-b^2=(a-b)(a+b)$ is something that the asker will most likely apply in the future... +1 – 1015 Feb 6 '13 at 21:49
It's a stretch for most people to think of $x - 49$ as something that can be expressed in the form $a^2 - b^2$. – Paul Raff Feb 6 '13 at 21:50
Well @PaulRaff , perhaps it shouldn't when most people are trying to deal with mathematics. I think this and other answers really remark the importance of knowing this little formulae for future applications in mathematics, and extending them to some not-so-obvious cases , like taking the square roots of $\,x\,$ – DonAntonio Feb 6 '13 at 23:16

Let $\sqrt{x}=y$. We are then looking at $\displaystyle\lim_{y\to 7}\frac{y-7}{y^2-49}$. Familiar?

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$$\begin{align} \lim_{x\to 49}\frac{\sqrt{x}-7}{x-49}=&\lim_{x \to 49}\frac{(\sqrt{x}-7)}{x-49}\frac{(\sqrt{x}+7)}{(\sqrt{x}+7)}\\ &=\lim_{x \to 49}\frac{(\sqrt{x})^2-7^2}{x-49}\frac{1}{(\sqrt{x}+7)}\\ &=\lim_{x \to 49}\frac{x-49}{x-49}\frac{1}{(\sqrt{x}+7)}\\ &=\lim_{x \to 49}\frac{1}{\sqrt{x}+7}\\ &=\frac{1}{14} \end{align}$$

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If you've learned L'Hôpital's rule, you know that this reduces to:

$$\lim_{x\to49}\frac{1}{2\sqrt{x}} = \frac{1}{14}$$

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We know that the difference between two squares is $x^2-a^2=(x-a)(x+a)$ so by writting $x-49$ as a difference between two squares we get$$x-49=(\sqrt{x}^2-7^2)=(\sqrt{x}-7)(\sqrt{x}+7)$$ which implies that $$\lim_{x\rightarrow 49} \frac{\sqrt{x}-7}{(\sqrt{x}-7)(\sqrt{x}+7)}=\lim_{x\rightarrow 49} \frac{1}{(\sqrt{x}+7)}=\frac{1}{14}.$$

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