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Limit problem

  1. Oct 12, 2005 #1
    Hey what's up?

    The teacher gave us a sheet with questions in order to prepare for the test which is today. There's 1 problem which i've tried doing a couple of times but I always get blocked at the same place.

    Evaluate the following limit by the Algebraic Method:

    lim(x->-4) (5x-4)/(2x^2+7x-4)



    I tried factoring the denominator which gives (2x-1)(x+4) but that doesn't help
    I tried replacing X by -4 but that's a division by 0 so i can't do it that way.

    Can anyone help me out with this one?
    Thanks in advance
     
  2. jcsd
  3. Oct 12, 2005 #2

    arildno

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    So, you are to evaluate: [tex]\lim_{x\to-4}\frac{5x-4}{(2x-1)(x+4)}[/tex]
    Do you agree that if you pick a number slightly bigger than x=-4, then you can make the fraction into an arbitrarily large positive number just by picking your number close enough to x=-4?
    Do you agree that if you pick a number slightly less than x=-4, then you can make the fraction into an arbitrarily "large" negative number just by picking your number close enough to x=-4?

    Can there exist a limit value (a NUMBER!) to which the fraction approaches when x goes to -4, or must we conclude that the limit doesn't exist at x=-4?
     
  4. Oct 12, 2005 #3
    Hey thanks man, yeah i can't believe i forgot about the LHL and RHL. *slaps self*

    Can I ask you 1 more question man? It's the last question on my work sheet.

    I know that in square root functions the numerator must be rationalised and that i must divide the whole function by negative square root of X squared when the limit is negative infinity but in this case the highest degree of the denominator is a half. How can i work this one out?

    lim (x-> negative infinity) (2x+1)/sqrt(x+1)

    Thanks a bunch
     
    Last edited: Oct 12, 2005
  5. Oct 12, 2005 #4

    TD

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    I assume you're working real (ignore this if you're working complex), what will the expression in the square root become? Is that defined in the reals?
     
  6. Oct 12, 2005 #5
    yeah i'm working in the reals

    the expression in the square root would be negative infinity. I don't know how to work this one out :(
     
  7. Oct 12, 2005 #6

    TD

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    What do you know about square roots of negative numbers in the reals (although negative infinity isn't really a number here, but we're approaching it so just ignore that subtle remark)?

    What is, for example, [itex]\sqrt{-1}[/itex] ?
     
  8. Oct 12, 2005 #7
    yeah we haven't seen that yet.. can't do the square root of a neg. number
     
  9. Oct 12, 2005 #8

    TD

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    You didn't see it because it "doesn't exist", i.e. it isn't defined for the reals.
    Look at the limit as a function, the negative values for x (x < -1 that is) aren't in the domain so the function isn't defined there.
     
  10. Oct 12, 2005 #9
    I understand that through direct substitution it wouldn't be possible as you can't find the square root of a neg. number but usually there are workaround for these types of problems, like multiply the whole fraction by either the nominator or denominator's congugate.

    How would you find it if it were positive infinity? You wouldn't just substitute X by infinity you'd have to rationalize the denominator.

    I hate square root functions! :cry:
     
  11. Oct 12, 2005 #10

    TD

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    For positive infinity, the limit would be (positive) infinity as well since rougly said; the degree of x in the nominator (1) is larger than the one of the denominator (1/2).

    For the negative infinity though, there won't be any 'tricks' since the problem isn't that you get an indeterminate form or something like that, but simply the fact that the function isn't even defined there and not even in a neighbourhood of the 'point'.

    For example, what would the following limit be according to you:

    [tex]\mathop {\lim }\limits_{x \to - 4} \sqrt x [/tex]

    Remember, we're working in the reals.

    Possibly, go back to the intrinsic definition of a limit and see if that helps you.
     
    Last edited: Oct 12, 2005
  12. Oct 12, 2005 #11
    The limit doesn't exist in that fct but that's pretty simple i mean there's only one term.

    Here's an example of what i mean

    lim (x-> negative infinity) (sqrt(x^2+x+1)+x)
    The final answer is -1/2

    But you would think that through direct substitution it wouldn't be possible since the square root of neg. infinity is infinity then you'd have to substract infinity to that but that's just not possible, and how could you then extract the square root from that...

    But when you multiply this function by 1 (congugate/congugate) you can work your way to the answer. This is what i'm looking for.
     
  13. Oct 12, 2005 #12
  14. Oct 12, 2005 #13

    TD

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    Ah but that's different! "Thanks to" the square within the root, this function actually is defined for negative numbers. That is not the case for your function! Look at the definition, for a limit to exist, you need to have a neighbourhood where the images of your arguments (when your argument goes to a certain x-value) approach your limit, arbitrary close. Is that the case here?
     
  15. Oct 12, 2005 #14
    Oh ok i understand what you mean now. I thought there would be a way to eliminate the square root on the denominator but i guess you just have to calculate x+1=0 from the start to find the domain of the function. I didn't know that with limits. All this factorizing 0/0 functions makes me feels like everything is possible
     
    Last edited: Oct 12, 2005
  16. Oct 12, 2005 #15

    TD

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    If you've seen the epsilon-delta definition, you'll see why it's useless to take a limit to a certain point a if the function isn't defined in at least a neighbourhood of that a. Moreover, for a limit of x going to a, x has to be in the domain of your function, for all x.
     
  17. Oct 12, 2005 #16
    Hey TD, if i told you i found you seductive, would you hold it against me?



    Just kidding thanks for the help bro.
     
  18. Oct 12, 2005 #17

    TD

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    You're welcome, but someone else may respond with some further information if I wasn't entirely correct. As far as I can tell though, this limit does not exist :smile:
     
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