# Math Help - Another limit question.

1. ## Another limit question.

I got another question that was lim as x->3 (3x-8)/(4x-12). I didn't see if it was possible to factor anything so I found that the limit doesn't exist because the left hand limit doesn't equal the right hand limit. for this question I just wanted to know if you could have found another answer algebraically.

2. Originally Posted by darksoulzero
I got another question that was lim as x->3 (3x-8)/(4x-12). I didn't see if it was possible to factor anything so I found that the limit doesn't exist because the left hand limit doesn't equal the right hand limit. for this question I just wanted to know if you could have found another answer algebraically.
$\lim_{x \to 3}\frac{3x - 8}{4x - 12} = \lim_{x \to 3}\frac{3x - 8}{\frac{4}{3}(3x - 9)}$

$= \frac{3}{4}\lim_{x \to 3}\frac{3x - 8}{3x - 9}$

$= \frac{3}{4}\lim_{x \to 3}\frac{3x - 9 + 1}{3x - 9}$

$= \frac{3}{4}\lim_{x \to 3}\left(1 + \frac{1}{3x - 9}\right)$

$\to \infty$.

3. Originally Posted by Prove It
$\lim_{x \to 3}\frac{3x - 8}{4x - 12} = \lim_{x \to 3}\frac{3x - 8}{\frac{4}{3}(3x - 9)}$

$= \frac{3}{4}\lim_{x \to 3}\frac{3x - 8}{3x - 9}$

$= \frac{3}{4}\lim_{x \to 3}\frac{3x - 9 + 1}{3x - 9}$

$= \frac{3}{4}\lim_{x \to 3}\left(1 + \frac{1}{3x - 9}\right)$

$\to \infty$.
I'm not really sure but if the limit goes to infinity does it exist? I'm not sure what your answer means.

4. Originally Posted by Prove It
$\lim_{x \to 3}\frac{3x - 8}{4x - 12} = \lim_{x \to 3}\frac{3x - 8}{\frac{4}{3}(3x - 9)}$

$= \frac{3}{4}\lim_{x \to 3}\frac{3x - 8}{3x - 9}$

$= \frac{3}{4}\lim_{x \to 3}\frac{3x - 9 + 1}{3x - 9}$

$= \frac{3}{4}\lim_{x \to 3}\left(1 + \frac{1}{3x - 9}\right)$

$\to \infty$.
Since $\displaystyle {\lim_{x \to 3^+}\frac{3x - 8}{4x - 12} = + \infty}$ and $\displaystyle {\lim_{x \to 3^-}\frac{3x - 8}{4x - 12} = - \infty}$ the limit does not exist.

Note that $f(x) = 1 + \frac{1}{3x - 9}$ is a hyperbola. It has a vertical asymptote at x = 3 and the above limits are exactly what you see when you draw its graph.

5. I think I've made a mistake here...

If the function approaches $3$ from the left, then we have the denominator with negative values, so approaching $-\infty$, while if the function approaches $3$ from the right, then we have the denominator with positive values, so approaching $\infty$.

Therefore the function does not exist.

6. Thanks for your help; I was able to find the answer by evaluating the left and right limit, but I wasn't sure how to do it algebraically.

7. "Trying" to set x= 3 and observing that you get " $\frac{1}{0}$" is sufficient to conclude that the limit does not exist. Of course, saying that the limit is $\infty$ is just another way of saying there is no limit.

8. Originally Posted by HallsofIvy
"Trying" to set x= 3 and observing that you get " $\frac{1}{0}$" is sufficient to conclude that the limit does not exist. Of course, saying that the limit is $\infty$ is just another way of saying there is no limit.
I would argue there's significant difference between something like $\displaystyle{ \lim_{x \to 1} \frac{1}{(x-1)^2} = + \infty}$ on the one hand and $\displaystyle{ \lim_{x \to 1} \frac{1}{x-1}}$ which does not exist (and in particular, is not equal to $\infty$) on the other hand.

9. Originally Posted by mr fantastic
I would argue there's significant difference between something like $\displaystyle{ \lim_{x \to 1} \frac{1}{(x-1)^2} = + \infty}$ on the one hand and $\displaystyle{ \lim_{x \to 1} \frac{1}{x-1}}$ which does not exist (and in particular, is not equal to $\infty$) on the other hand.
I agree, it depends on the context in which the symbol $\infty$ is being used.

10. Originally Posted by mr fantastic
I would argue there's significant difference between something like $\displaystyle{ \lim_{x \to 1} \frac{1}{(x-1)^2} = + \infty}$ on the one hand and $\displaystyle{ \lim_{x \to 1} \frac{1}{x-1}}$ which does not exist (and in particular, is not equal to $\infty$) on the other hand.
Yes, it is saying that the limit does not exist, for a specific reason and so is more precise than just saying "the limit does not exist". But it is still saying "the limit does not exist".