# How do you find the limit of sqrt(x-3)/(x-9) as x approaches 9?

Jul 15, 2015

lim_{xto9}sqrt(x-3)/(x-9) = ∞

#### Explanation:

Let's try direct substitution.

$\frac{\sqrt{x - 3}}{x - 9} = \frac{\sqrt{9 - 3}}{9 - 9} = \frac{\sqrt{6}}{0}$

So there is a "hole" at $x = 9$.

lim_{xto9}sqrt(x-3)/(x-9) = lim_{xto9}sqrt6/0 = ∞

Jul 15, 2015

The answer you are expected to write depends, in part, on where and with whom you are studying Calculus.

#### Explanation:

In James Stewart's Calculus (used at many college and universities in the US), the final answer to this question is "Does Not Exist".

As $x \rightarrow {9}^{+}$ (as $x$ approaches $9$ from the right), we get $\sqrt{x - 9} \rightarrow \sqrt{6}$, a positive limit and $x - 9 \rightarrow {0}^{+}$ That is, the numerator approaches a positive limit and thedenominator approaches $0$, through positive values.

As $x \rightarrow {9}^{+}$, the quotient: $\frac{\sqrt{x - 3}}{x - 9}$ increases without bound.
We write: ${\lim}_{x \rightarrow {9}^{+}} \frac{\sqrt{x - 3}}{x - 9} = \infty$

Coming from the left, we still get the numerator approaching a positive limit, by now, as $x \rightarrow {9}^{+}$, we get $x - 9 \rightarrow {0}^{-}$ That is, the numerator approaches a positive limit and thedenominator approaches $0$, through negative values.

As $x \rightarrow {9}^{-}$, the quotient: $\frac{\sqrt{x - 3}}{x - 9}$ decreases without bound.
We write: ${\lim}_{x \rightarrow {9}^{-}} \frac{\sqrt{x - 3}}{x - 9} = - \infty$

In this case, our Stewart (and others) adopt the convention that the limit (that is, the "two-sided" limit) simply does not exist.

We write: lim_(xrarr9^)sqrt(x-3)/(x-9) " Does not exist"

Of course, "infinite limits" do not exist either, but when we write, for example: ${\lim}_{x \rightarrow 0} \frac{1}{x} ^ 2 = \infty$, we are saying why the limit does not exist. (Because the quotient increases without bound.)