# Evaluating the limit of $\lim_{x \to 9} \frac {x-9} {\sqrt{x} - 3}$

How do you evaluate the limit of $$\lim_{x \to 9} \frac {x-9} {\sqrt{x} - 3}$$

by first principle? I was thinking of this way to evaluate but it seems to evaluate to $\frac {0} {0}$.

$$\lim_{x \to 9} \frac {x-9} {\sqrt{x} - 3} = \frac{\lim_{x \to 9} (x-9) }{\lim_{x \to 9} {\sqrt{x} - 3}}$$

I mentioned that the limits do not exist because the function is not continuous at $x=9$. Any idea on how to solve this question?

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Hint: Can you factor the numerator in $(\sqrt x \pm 3)$ terms and then simplify? You could also do a series expansion or use first principles, but the above hint is easiest. Regards – Amzoti Feb 1 '13 at 16:49
Let $x=y^2$ ($y\ge 0$). We are looking at $\frac{y^2-9}{y-3}$. Familiar-looking? – André Nicolas Feb 1 '13 at 18:09

$$\frac{x-9}{\sqrt x-3}=\frac{(x-9)(\sqrt x+3)}{(x-9)}=\sqrt x+3\text{ if } x\ne 9$$

Now, if $x\to 9,x\ne 9$

So, $$\lim_{x\to 9}\frac{x-9}{\sqrt x-3}=\lim_{x\to 9}(\sqrt x+3)=\sqrt 9+3=6$$

Alternatively, let's consider $$\lim_{x\to a^2}\frac{x-a^2}{\sqrt x-a}$$

If we put $h=\sqrt x-a$ so that $h\to0$ as $x\to a^2$ and $x=(a+h)^2$

So, $$\lim_{x\to a^2}\frac{x-a^2}{\sqrt x-a}=\lim_{h\to0}\frac{(a+h)^2-a^2}h=2a \text{ as }h\to0\implies h\ne0$$

Here $a=3$

Also observe that $$\lim_{h\to0}\frac{(a+h)^2-a^2}h=\frac{d (x^2)}{dx}_{(\text{ at }x=a) }$$
by the Differentiation from First Principles or Definition via difference quotients.

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Multiply the expression's numerator and denominator by the conjugate of $\;\sqrt x - 3 = \sqrt x + 3$

$$\frac{x-9}{\sqrt x-3}=\frac{(x-9)(\sqrt x + 3)}{(\sqrt x - 3)(\sqrt x + 3)} = \frac{(x-9)(\sqrt x+3)}{(x-9)}=\sqrt x+3\quad\text{ provided }\; x\ne 9$$

Note: we evaluate the limit of $f(x)$ as $x\to 9$; we are not evaluating $f(9)$. So there is no problem if it happens to be the case that the expression is valid for $x\ne 9$.

Recall, evaluating a limit is determining the value of the expression as $x \to 9$, not finding its value AT $x = 9$. So the expression may be discontinuous AT 9, but its limit is the value we obtain when $x$ gets VERY CLOSE to 9.

This gives us:

$$\lim_{x\to 9}\frac{x-9}{\sqrt x-3}=\lim_{x\to 9}(\sqrt x+3)=\sqrt 9+3=6$$

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Does this make sense, bryansis2010? – amWhy Feb 1 '13 at 21:12

$$\lim_{x \rightarrow 9} \frac{x-9}{\sqrt(x)-3} = \lim_{x \rightarrow 9} \frac{1}{\frac{1}{2\sqrt{x}}} = 6$$

Here L'Hopital's rule is used.

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appreciate this because i'm learning L'Hopital rule too – bryansis2010 Feb 1 '13 at 16:57

multiplied by $\frac{\sqrt{x}+3}{\sqrt{x}+3}$ or you can factorized it,remember $a^2-b^2=(a-b)(a+b)$ so you can get $x-9=(\sqrt{x}+3)(\sqrt{x}-3)$

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