Another limit question.

**darksoulzero** · July 9th 2010, 10:56 PM

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.

**Prove It** · July 10th 2010, 12:41 AM

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.

$\lim_{x \to 3}\frac{3x - 8}{4x - 12} = \lim_{x \to 3}\frac{3x - 8}{\frac{4}{3}(3x - 9)}$

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

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

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

$\to \infty$ .

**darksoulzero** · July 10th 2010, 01:50 AM

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.

**mr fantastic** · July 10th 2010, 01:54 AM

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.

**Prove It** · July 10th 2010, 01:55 AM

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.

**darksoulzero** · July 10th 2010, 02:51 AM

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.

**HallsofIvy** · July 10th 2010, 06:32 AM

"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.

**mr fantastic** · July 10th 2010, 03:44 PM

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.

**Prove It** · July 10th 2010, 07:49 PM

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.

**HallsofIvy** · July 11th 2010, 02:19 AM

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".

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