Question

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

Answer 1

`lim_{xto9}sqrt(x-3)/(x-9) = ∞`

Explanation:

Let's try direct substitution.

`sqrt(x-3)/(x-9) = sqrt(9-3)/(9-9) = sqrt6/0`

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

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

Answer 2

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 `xrarr9^+` (as `x` approaches `9` from the right), we get `sqrt(x-9) rarr sqrt6`, a positive limit and `x-9 rarr 0^+` That is, the numerator approaches a positive limit and thedenominator approaches `0`, through positive values.

As `xrarr9^+`, the quotient: `sqrt(x-3)/(x-9)` increases without bound.
We write: `lim_(xrarr9^+)sqrt(x-3)/(x-9) = oo`

Coming from the left, we still get the numerator approaching a positive limit, by now, as `xrarr9^+`, we get `x-9 rarr 0^-` That is, the numerator approaches a positive limit and thedenominator approaches `0`, through negative values.

As `xrarr9^-`, the quotient: `sqrt(x-3)/(x-9)` decreases without bound.
We write: `lim_(xrarr9^-)sqrt(x-3)/(x-9) = -oo`

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_(xrarr0)1/x^2 = oo`, we are saying why the limit does not exist. (Because the quotient increases without bound.)