1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding Hard Limits

  1. Aug 7, 2008 #1
    I started learning limits and I have some difficulties qith this question;
    Find the limits or say why they do not exist.

    a)[tex]lim_{x\rightarrow2}\frac{\left|x^2 + x -6\right|}{x-2}[/tex]

    b) [tex]lim_{x\rightarrow0}\frac{\sqrt{4-x}-2}{\sqrt{x+1}-1}[/tex]




    3. The attempt at a solution

    a)we have an absolute value on the numerator but we can factorise it:
    [tex]lim_{x\rightarrow2}\frac{\left|x^2 + x -6\right|}{x-2}[/tex]
    [tex]lim_{x\rightarrow2}\frac{(x-2)(x+3)}{x-2}[/tex]

    Now we have two cases:

    (x approaching 2 from the positive side)
    [tex]lim_{x\rightarrow2+}\frac{(x-2)(x+3)}{x-2}[/tex]

    [tex]lim_{x\rightarrow2+}\frac{(2-2)(2+3)}{2-2}[/tex] = undefined (?)

    (x approaching 2 from the negative side)
    [tex]lim_{x\rightarrow2-}\frac{(-x+2)(x+3)}{x-2}[/tex]

    [tex]lim_{x\rightarrow2-}\frac{(-2+2)(2+3)}{2-2}[/tex] = undefined

    Why do I always get devision by zero and undefined, did I do somethin wrong?


    b) [tex]lim_{x\rightarrow0}\frac{\sqrt{4-x}-2}{\sqrt{x+1}-1}[/tex]

    Because we have a sqrt on the denominator we mulitply top & bottom by the conjugate and we rationalize the denominator,

    [tex]lim_{x\rightarrow0}\frac{(\sqrt{4-x}-2)(\sqrt{x+1}+1)}{(\sqrt{x+1}-1)\sqrt{x+1}+1)}[/tex]

    The denominator expands and simplifies to:

    [tex]lim_{x\rightarrow0}\frac{(\sqrt{4-x}-2)(\sqrt{x+1}+1)}{x+1-1}[/tex]

    What can i do next? should I expand the numerator too? ( it would become
    [tex]\sqrt{4-x.x+1}+\sqrt{4-x}-2\sqrt{x+1}-2[/tex])
    But I don't know how to do this one.

     
  2. jcsd
  3. Aug 7, 2008 #2
    1st question :
    you are right to split the limit into 2 cases
    it's easier if in the second case you factor out the -1from the brackets and keep the factors the same as in the first case (x-2)
    What you are forgetting to do is cancel the factors from the numerator and the denominator...once you cancel an (x-2) , the limits are no longer undefined.... so cancel it in both cases, and compare the 2 answers, and if they are the same, the limit exists
     
  4. Aug 7, 2008 #3
    For part a), you would have two cases as you stated above. The (x-2) term on the top and bottom should "cancel" each other out leaving you with an exact value. You must check if the limit approaching from the right matches the limit approaching from the left to see if the limit exists. You're on the right track.

    Now for part b), you are almost there. You managed to rationalize the denominator but there is still work to be done with the numerator. No expansion is required. Think about what else can be done!!! The (sqrt(4-x) - 2) term is giving you a problem right ( 0/0 )?
     
  5. Aug 7, 2008 #4
  6. Aug 7, 2008 #5
    Are we allowed to cancel out?


    We haven't studied L'Hôpital's rule but I don't know what to do with that one..
     
  7. Aug 7, 2008 #6
    btw, did i use the correct expression in a, in the numerator, when x approaching 2 from the negative side?

     
  8. Aug 7, 2008 #7
    No, you're right. You're able to cancel. If the numerator and denominator has two common terms (in this case, (x-2)), it is legitimate to cancel out.

    Case 1.) Positive Limit (lim x->2+)

    [tex]lim_{x\rightarrow2+}\frac{(x-2)(x+3)}{x-2}[/tex]

    Since (x-2) is a common factor... you can cancel it out. Remember, this makes a 1:

    [tex]lim_{x\rightarrow2+}\frac{(1)(x+3)}{1}[/tex]

    Substitute x = 2 into the above case, and this is your right hand limit.

    Case 2.)

    [tex]lim_{x\rightarrow2-}\frac{(-x+2)(x+3)}{x-2}[/tex]

    I'm not holding your hand on this one. Do the same thing but hint: factor out a -1 to have a common term.
     
  9. Aug 7, 2008 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    One "limit rule" that doesn't get as much notice as it deserves is this:

    "If g(x)= f(x) for all x except x= a, then [itex]\lim_{x\rightarrow a} f(x)= \lim_{\rightarrow a} g(x)[/itex]."

    In this particular case,
    [tex]\frac{x^2+ x- 6}{x-2}= \frac{(x-2)(x+3)}{x-2}[/tex]
    and
    [tex]\frac{(x-2)(x+3)}{x-2}= x+3[/tex]
    as long as x is not equal to 2. Therefore, they have the same limit. Yes, in a limit situation such as this you can "cancel" the 0s.

    As for
    [tex]lim_{x\rightarrow 0}\frac{\sqrt{4-x}-2}{\sqrt{x+1}-1}[/tex]
    While you can use L'Hopital's rule, often that is "over kill". With a problem like this you should think first of "rationalizing" both the numerator and the denominator: Multiply both numerator and denominator by [itex](\sqrt{x+1}+ 1)(\sqrt{4-x}+2)[/itex]:
    [tex]\frac{\sqrt{4- x}- 2}{\sqrt{x+1}-1}= \frac{(\sqrt{4-x}-2)(\sqrt{4-x}+2)(\sqrt{x+1}+1)}{(\sqrt{x+1}-1)(\sqrt{x+1}+1)(\sqrt{4- x}+ 2)}[/tex]
    [tex]= \frac{(4-x-4)(\sqrt{x+1}+1)}{\sqrt{4-x}+ 2)(x+ 1- 1)}= \frac{x(\sqrt{x+1}+1)}{x(\sqrt{4-x}+ 2)}= \frac{\sqrt{x+1}+1}{\sqrt{4-x}+ 2}[/tex]
    Now, what is the limit of that as x goes to 0?
     
    Last edited: Aug 8, 2008
  10. Aug 7, 2008 #9
    Leave it as:

    [tex]lim_{x\rightarrow0}\frac{(\sqrt{4-x}-2)(\sqrt{x+1}+1)}{x+1-1}[/tex]

    But simply it so the denominator is just x.


    So we still can't substitute, because it will still yield an indeterminate form (0 / 0). I worked it out a few minutes ago, and you can solve this WITHOUT L'Hopital's Rule (Anyway...assuming this is Calc 1... you haven't even learned Derivatives yet).

    The fundamental problem in most hard limit problems is: "How can I remove that pesky x so we're not dividing by zero?"

    Here's a hint: ( sqrt{4-x}-2 ) * ( sqrt{4-x} + 2 ) = 4 - x - 4 = -x . That looks real delicious.
    Multiply by the conjugate... again. This time with the numerator. Tell me what you get. (The answer is not zero FYI)
     
  11. Aug 7, 2008 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I once taught a "Calculus for Economics and Business Administration Majors" course that use an awful book!

    On one page they listed the laws of limits:
    If lim f= A and lim g= B then lim f+ g= A+ B.
    If lim f= A and lim g= B then lim fg= AB
    If lim f= A and lim g= B and [itex]B\ne 0[/itex], then lim f/g= A/B.

    On the very next page they introduced the integral- completely ignoring the fact that in that very important case, those laws don't apply! The derivative is the limit of the fraction f(x+h)- f(x)/h as h goes to 0 and so the "B" above must be 0. What was missing was precisely the rule I mention above.
     
  12. Aug 7, 2008 #11

    HallsOfIvy,

    I do not quite understand this limit rule (something that I haven't turned upon on my studies). Is there an external source where I can study this?

    Also, in your explanation for Question #1...are you saying that the limits are the same?

    Carlo
     
    Last edited: Aug 7, 2008
  13. Aug 7, 2008 #12
    The limit is 1. Am I right?
     
  14. Aug 7, 2008 #13
    Uh, check your arithmetic.
     
  15. Aug 7, 2008 #14

    By the way: the end product should be -(sqrt(x+1) + 1) / (sqrt(4-x) + 2).

    The negative sign of x was forgotten when the numerator was 4 - x - 4.

    Carlo
     
  16. Aug 8, 2008 #15

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Do you mean:
    You should have learned that along with the definition of limit- but, as I said, students often don't notice it.

    The Definition of "limit" is this:
    "The [itex]\lim_{x\rightarrow a} f(x)= L[/itex]" means that, given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]0< |x- a|< \delta[/itex] then [itex]|f(x)- L|< \epsilon[/itex].

    Notice that "0< |x-a|"? What happens at x= a is irrelevant to the limit! That means that we can change the value of the function in any way we want at x= a and not change the limit there.

    Here's a cute problem: Define f(x) by
    f(x)= 1000 if x< -0.001
    f(x)= 3x2+ 6x if [itex]-0.001\le x< 0
    f(x)= 1000 if x= 0
    f(x)= sin(x) if [itex]0< x\le 0.001
    f(x)= 1000 if x> 0.001

    What is [itex]\lim_{x\rightarrrow 0} f(x)[/itex]?
     
  17. Aug 8, 2008 #16
    HallsOfIvy,

    Yes, I've learned the formal definition of a limit (Epsilon-Delta), and I understand that even though there might be discontinuities at x = a (as the limit of x approaches a), there is no bearing on the actual limit. (that is... the limit as x approaches a still exists even if there is still a point discontinuity).

    1.) The "one limit rule that students mostly overlook," as you've stated, is kind of a blur. Perhaps the latex was mistyped on the last part of the section? Please correct me if I translate it wrong into English... but this is my interpretation of "the rule:"

    If g(x) = f(x) for all x except x = a, then the limit as x approaches a of f(x) is equal to the limit * a(g(x)).

    Perhaps the latter half of the sentence means, "the limit of x approaches to a is equivalent to g(x). If this is so, then this rule isn't a problem anymore--I just misread it.

    One last thing--what bearing does the Epsilon-Delta rule have on this problem (Problem #1), when #1's answer is: The Limit DNE?

    Carlo
     
  18. Aug 8, 2008 #17

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I hadn't noticed that I had the LaTex wrong. It should have been
    [itex]\lim_{x\rightarrow a} f(x)= \lim_{\rightarrow a} g(x)[/itex]


    By "#1" do you mean "a":
    [tex]\lim_{x\rightarrow2}\frac{\left|x^2 + x -6\right|}{x-2}[/tex]

    It has already been shown that, as x goes to 2 from above, the "one-sided" limit is (2+ 3)= 5 and, as x goes to 2 from below, the "one-sided" limit is -(2+ 3)= -5 so the limit itself does not exist.

    What does that have to do with "epsilon-delta"? That shows that for any [itex]\delta> 0[/itex] we can find x closer to 2 than that ([itex]|x- 2|< \delta[/itex]) with x> 2 arbitrarily close to 5 and with x< 2, arbitrarily close to -5. If we were to claim that the limit is any number, L, then we would have to have [itex]|f(x)- L|= |5- L|< \epsilon[/itex] and [itex]|f(x)- L|= |-5-L|= |5+L|< \epsilon[/itex] for values of x within the same [itex]\delta[/itex] of 2. It should be clear that those can't both be true for any [itex]\epsilon< 5[/itex].
     
  19. Aug 8, 2008 #18
    Thank you! That cleared up some of my thoughts.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Finding Hard Limits
  1. Hard Limit Problem (Replies: 5)

  2. A Hard Factorial Limit (Replies: 11)

  3. Really hard limits (Replies: 2)

Loading...