# limit question (mult. choice) do u agree?

• Jul 15th 2009, 07:09 AM
janedoe
limit question (mult. choice) do u agree?
http://i26.tinypic.com/216kqr.jpg

^It's all there

Thanks!
• Jul 15th 2009, 07:15 AM
Jester
I agree. You'll notice from the graph of $\displaystyle f$, $\displaystyle f$ is negative near $\displaystyle x = -3$ and $\displaystyle x = 3$ so b) can't be true.
• Jul 15th 2009, 07:17 AM
Rapha
Hello!

You are right, a) and c) (that is answer d :-) is correct)

Of course b is not correct.

You know that answer a) is correct and probably you know that the solution is -7, but what matters is that $\displaystyle \lim_{x \to -3+ }f(x) = \lim_{x \to -3- }f(x)$ (thats answer a) )

Using that answer, how is answer b) : $\displaystyle \lim_{x \to -3+ }f(x) = -\lim_{x \to -3- }f(x)$right?

That's nuts (Wink) ; assuming b) is correct, it is $\displaystyle \lim_{x \to -3+ }f(x) = -\lim_{x \to -3- }f(x) = \lim_{x \to -3- }f(x)$, e. g. $\displaystyle -\lim_{x \to -3- }f(x) = \lim_{x \to -3- }f(x)$ : If the limes of f(x) is '0' it would have been correct : But it's not!

Yours
Rapha
• Jul 15th 2009, 07:18 AM
HallsofIvy
L is not equal to -L! (as long as L is not 0)

In particular, -7 is not equal to 7.

(a) and (c) really say the same thing: If X= Y then Y= X. But (b) has that "-" in it.
• Jul 15th 2009, 07:32 AM
janedoe
Quote:

Originally Posted by HallsofIvy
L is not equal to -L! (as long as L is not 0)

In particular, -7 is not equal to 7.

(a) and (c) really say the same thing: If X= Y then Y= X. But (b) has that "-" in it.

Well, I know 7 and -7 are not equal, but that's why I thought 7 and -(-7) which is 7 would be. What does that negative sign in front of the limit in b do then?
• Jul 15th 2009, 07:39 AM
Rapha
Hi again.

Quote:

Originally Posted by janedoe
Well, I know 7 and -7 are not equal, but that's why I thought 7 and -(-7) which is 7 would be. What does that negative sign in front of the limit in b do then?

Just consider the solutions of answer a,b,c

a)
$\displaystyle -7=\lim_{x\to -3^+} f(x)= \lim_{x\to -3^-}f(x)=-7$

b)

$\displaystyle -7 = \lim_{x\to -3^-} f(x)\not= -\lim_{x\to -3^+} f(x) = -(-7) = +7$

You still have a -7 on the left side.

c)
$\displaystyle -7= \lim_{x\to -3^-} f(x) = \lim_{x\to -3^+}f(x) =-7$
• Jul 15th 2009, 09:04 AM
janedoe
I'm sorry but I'm still not getting your logic..I'm not sure why you changed most of the signs in each answer choice. I'm not sure if you did it by accident or on purpose, but I don't get why.

For example, in choice b, it's:
x-->3- and then on the other side, x-->3+
I still can't see why my reasoning is wrong... +7 = -(-7)

You put: -3- and -3+

For c:
-3- and 3+ (but you put -3+)
• Jul 15th 2009, 09:21 AM
Plato
Quote:

Originally Posted by janedoe
For example, in choice b, it's:
x-->3- and then on the other side, x-->3+
I still can't see why my reasoning is wrong... +7 = -(-7)
You put: -3- and -3+
For c:
-3- and 3+ (but you put -3+)

$\displaystyle \lim _{x \to - 3^ - } \frac{7} {{8 - x^2 }} = \lim _{x \to - 3^ + } \frac{7} {{8 - x^2 }} = \lim _{x \to 3^ - } \frac{7} {{8 - x^2 }} = \lim _{x \to 3^ + } \frac{7} {{8 - x^2 }} = - 7$

Do you understand why all those limits equal $\displaystyle -7$?
• Jul 15th 2009, 09:33 AM
janedoe
Quote:

Originally Posted by Plato
$\displaystyle \lim _{x \to - 3^ - } \frac{7} {{8 - x^2 }} = \lim _{x \to - 3^ + } \frac{7} {{8 - x^2 }} = \lim _{x \to 3^ - } \frac{7} {{8 - x^2 }} = \lim _{x \to 3^ + } \frac{7} {{8 - x^2 }} = - 7$

Do you understand why all those limits equal $\displaystyle -7$?

I understand why when you actually plug in -3 or 3, but I was getting confused because I was also viewing the graph, and thinking that the big u shape in the middle was also approaching positive 7 but I guess not. I thought I had to use the graph to see what its approaching from the left or right side. Can I just do it mathematically?
• Jul 15th 2009, 09:54 AM
Plato
Quote:

Originally Posted by janedoe
Can I just do it mathematically?

I don't know any other ways to do it but mathematically.
Be sure that you have done the graph correctly (mathematically).
• Jul 15th 2009, 03:29 PM
janedoe
I guess I meant, plugging in numbers versus typing the function into the calculator and just figuring it out based on the graph. But thanks all, it makes sense now that I just did it by hand.