# Find the limit if it exists

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• September 12th 2012, 06:48 PM
calculus123
Find the limit if it exists
I'm having a really difficult time understanding limits from the graph. I don't completely understand how to find if the limit exists or not. I tried to do these for my homework but I don't know if they are right. Could someone please help me understand limits?
http://i45.tinypic.com/152hn36.jpg
• September 12th 2012, 06:54 PM
SworD
Re: Find the limit if it exists
B and M are incorrect. Can you recheck those and see what the problem is?

Also, the graph is a bit unclear as x -> 4... is that supposed to be an asymptote, or do the curves actually stop at $\pm5$? If theres an asymptote, the one sided limits will be $\pm\infty$, the regular two-sided limit won't exist. But I think that's an asymptote, so F and G are wrong too.

To answer your question, a one sided limit will exist if the function approaches a particular value (or to infinity), and gets arbitrarily close as you get closer and closer, FROM that direction.

The usual two-sided limit will exist ONLY if the left-hand and right-hand limits exist and are equal.

The function will be continuous if the limit exists, AND the limit actually equals the function value. Notice that this does actually happen at x=6.

Edit: A looks incorrect as well.
• September 12th 2012, 06:56 PM
MarkFL
Re: Find the limit if it exists
By my count, you have 4 incorrect.
• September 12th 2012, 07:14 PM
Soroban
Re: Find the limit if it exists
Hello, calculus123!

Most of your answers are correct . . . good work!

Quote:

I'm having a really difficult time understanding limits from the graph.
I don't completely understand how to find if the limit exists or not.
I tried to do these for my homework but I don't know if they are right.
http://i45.tinypic.com/152hn36.jpg

Your answer for (a) is incorrect.

Trace the graph approaching -3 from the left.
. . The graph approaches $f(x) =1.$

Trace the graph approaching -3 from the right.
. . The graph approaches $f(x) = 1.$

Therefore: . $\lim_{x\to\text{-}3}f(x) \;=\;1$

Your answer for (g) is off.
It looks like it goes to 5,
. . but I believe there is a vertical asymptote at $x = 4.$

Therefore: . $\lim_{x\to4^+}}f(x) \:=\:+\infty \quad\text{ and }\,\lim_{x\to4^-}f(x) \:=\:-\infty$

Your answer to (m) is incorrect.
$f(x)$ is continuous at $x = 6.$
(The graph doesn't "break" there, does it?)
• September 12th 2012, 08:01 PM
calculus123
Re: Find the limit if it exists
Thank you Soroban and SworD! I seem to understand them a little bit better now. I made all the changes that you guys suggested, are they correct now? I tried to redo them and they seem to make more sense now. Thanks!
http://i50.tinypic.com/33k4ya1.jpg
• September 12th 2012, 08:08 PM
SworD
Re: Find the limit if it exists
Only B is still incorrect. Notice that they want the limit from the right.
• September 12th 2012, 08:32 PM
calculus123
Re: Find the limit if it exists
So the answer to b would be 2 since it's coming from the right?
• September 12th 2012, 08:36 PM
SworD
Re: Find the limit if it exists
Yes. Limits don't care at all about the value at that point.
• September 13th 2012, 02:56 AM
Plato
Re: Find the limit if it exists
Quote:

Originally Posted by SworD
Only B is still incorrect. Notice that they want the limit from the right.

That in not correct.

Quote:

Originally Posted by calculus123
So the answer to b would be 2 since it's coming from the right?

That in not correct.

b) ${\lim _{x \to {2^ + }}}f = 1$.
• September 13th 2012, 12:37 PM
calculus123
Re: Find the limit if it exists
For b), It's ${\lim _{x \to {-2^ + }}}f = 1$, not ${\lim _{x \to {2^ + }}}f = 1$. Are you sure the answer is 1 and not 2? I'm looking at the graph again and it seems like the answer is 2.
• September 13th 2012, 02:09 PM
Plato
Re: Find the limit if it exists
Quote:

Originally Posted by calculus123
For b), It's ${\lim _{x \to {-2^ + }}}f = 1$, not ${\lim _{x \to {2^ + }}}f = 1$. Are you sure the answer is 1 and not 2? I'm looking at the graph again and it seems like the answer is 2.

The image you posted is very hard to read.
You should try to make questions clear to the reader.
• September 13th 2012, 02:32 PM
SworD
Re: Find the limit if it exists
It's pretty clear that the number being approached in (b) is -2... unless they decided to make the -> arrow twice as big and have a wedge drawn on it.
• September 13th 2012, 02:42 PM
Plato
Re: Find the limit if it exists
Quote:

Originally Posted by SworD
It's pretty clear that the number being approached in (b) is -2... unless they decided to make the -> arrow twice as big and have a wedge drawn on it.

It may be different on different computer screens. On mine it appears $\to 2$ as opposed to $\to -2$.
I cannot understand why any serious user of this site does not learn to use LaTeX.
• September 13th 2012, 07:16 PM
calculus123
Re: Find the limit if it exists
Quote:

Originally Posted by Plato
It may be different on different computer screens. On mine it appears $\to 2$ as opposed to $\to -2$.
I cannot understand why any serious user of this site does not learn to use LaTeX.

I don't blame you for not seeing the negative sign because I have the sheet in front of me printed and I can't see it either. Thanks for your help!
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