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A few calculus questions

  1. Feb 24, 2005 #1
    A few calculus questions....

    Hey all. I have a few questions involving calculus. Any help would be much appreciated.

    1) [tex] f(x)= \left\{\begin{array}{cc}x^2,&\mbox{if} x \leq 0\\3x+1, &\mbox{if} 0<x<4\\12-x^2,&\mbox{if} x>4\end{array}\right [/tex]

    Find the following limits,if they exist:

    i) [tex]\lim_{x\rightarrow 2} 3x+1 [/tex]
    [tex] = 3(2) +1 [/tex]
    [tex] \lim_{x\rightarrow 2} 7 [/tex]

    ii) [tex] f(0) [/tex]
    replacing every equation with 0, i got [tex] x^2= 0, 3x+1= 1, 12-x^2= 12 [/tex] the answer i got was that the limit does not exist. Would this be true in this case?

    2) Find the slope of the tangent line to the curve [tex] y=x^3-x [/tex] at the point (-2,-6) using:

    i) [tex]\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h} [/tex]

    ii) [tex]\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a} [/tex]

    i) Using the first formula I got an answer of 11 so technically, using the second formula should give me the same the answer but I can't get it and I have to show work. Here is what I have so far:

    [tex] \lim_{x\rightarrow -2} \frac {x^3-x-((-2)^3-(-2))}{x+2} [/tex]
    [tex] \lim_{x\rightarrow -2} \frac {x^3-x+6}{x+2} [/tex]

    Not applying L'Hospital's Rule and solving this by factoring, I need to get a [tex] (x+2) [/tex] on the top so the top and bottm [tex] (x+2) [/tex] equal one. And then I can put in a -2 for x since you're not going to be dividing by zero. I hope I make sense and could anyone tell me if I'm doing something wrong? Thanks.
     
  2. jcsd
  3. Feb 24, 2005 #2

    arildno

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    Taking your last bit of factoring first:
    Yep, this will work. Use polynomial division.
     
  4. Feb 24, 2005 #3

    dextercioby

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    For ii),you need to replace (actually compute the limit) only on the top two braches,because the third's domain DOES NOT INCLUDE "0"...

    As for the last,u shouldn't be using L'H^ospital's rule,because:
    1.It's technically "more advanced"...
    2.It can be solved in a much more simple way,as Arildno suggested.

    Daniel.
     
  5. Feb 24, 2005 #4
    For 1ii) , I calculated the limit for [tex] x^2 [/tex] and [tex] 3x+1 [/tex] and I got the answer 0 and 3 respectively as the limit. Would this be the correct way to approach that question if it asked to find the limit at f(0)?
     
  6. Feb 25, 2005 #5

    dextercioby

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    Yes,that would be the correct way to approach the problem,but,unfortunately,the result is not correct.

    Daniel.

    P.S.You may wanna check again the rule of addition between 0 and 1.
     
  7. Feb 25, 2005 #6

    BobG

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    More specifically, the limit as you approach from the left
    [tex]\lim_{x\rightarrow 0^-} x^2 = 0 [/tex]

    The limit as you approach from the right
    [tex]\lim_{x\rightarrow 0^+} 3x+1 = 1[/tex]

    If the two limits aren't equal, then the limit
    [tex]\lim_{x\rightarrow 0} f(x) [/tex]
    does not exist.
     
  8. Feb 26, 2005 #7
    Got it. Thank you for all the help.
     
  9. Feb 26, 2005 #8

    Curious3141

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    I'm not clear on one thing ? Why does f(0) not exist ? There is a jump discontinuity at x = 0, but the function is still defined as [tex]f(x) = x^2[/tex] at x = 0, so f(0) = 0, isn't it ?

    The derivative at that point, of course, does not exist. But I think the function is still very much defined and its value is zero.
     
    Last edited: Feb 26, 2005
  10. Feb 26, 2005 #9

    BobG

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    f(0) does exist. The limit of f(0) doesn't exist.
     
  11. Feb 26, 2005 #10

    Curious3141

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    OK, that's clearer, thanks.
     
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