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Should the limit even exist?

  1. Jan 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Hello, I have some Calculus problems from my latest homework.

    The limit as x approaches 3 of sqrt(x+1)
    The back of the book says 2. But I'm wondering why it exists as sqrt(3+1) = sqrt(4) = +-2, so since there are 2 values for this, should the limit even exist?

    The limit of delta x as it approaches 0 of (sqrt(x + delta x) - sqrt(x)) divided by delta x.

    In similar problems, I factored the denominator out of the numerator so I could cancel the denominator so it could never go to 0. But I don't know how to do this for stuff in the radical sign. The book gives no example of this type of problem.

    3. Graphical, numerical, and analytical
    a. use a graphing utility to graph the function and estimate the limit.
    b. use a table to reinforce your conclusion
    c. find the limit by analytic methods.

    limit as x approaches 1 from the negative side of 2/(x^2-1).

    Since it's from the negative side, I just plugged 0.9999999 for x into my calculator, so the bottom approached 0 as a negative number. So I concluded that it is negative infinity. But isn't plugging in 0.999999 considered the numerical method? How do I do this analytically?

    2. Relevant equations

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 18, 2009 #2


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    Gold Member

    Re: limits

    Regarding one: The answer should not be +-2 as the sqrt(x+1) explicitly denotes the positive square root. Hence, if I asked you what sqrt(49) is equivalent to, you likely would not say +-7 but rather 7. However, given x^2 = 49 and asked to solve for x, you should give the answers +-7.

    Regarding two: I suggest that you multiply and divide by the conjugate of sqrt(x+deltax) - sqrt(x). sqrt(x+deltax) + sqrt(x) should be the conjugate.

    For number three: The only way I can think of doing it somewhat analytically would be to note that for x < 1 your expression is negative; therefore, as x tends to one from x < 1, f(x) -> -infinity
    Last edited: Jan 18, 2009
  4. Jan 18, 2009 #3


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    Homework Helper

    Re: limits

    This is a common misconception that results from a poor understanding of what is happening in, lets say, making x the subject in [tex]x^2=y[/tex]. Yes it will result in [tex]x=\pm \sqrt{y}[/tex] but because if you square either a positive or negative, you result in a positive. However, if you take [tex]\sqt{x}[/tex], x can ony be positive, not negative. So the limit here is only +2.

    Try multiplying through by the conjugate of the numerator :wink:
    Remember, [tex](x+y)(x-y)=x^2-y^2[/tex] so the radicals will be removed (in the numerator at least).

    I've never been taught how to find limits that tend to infinite by analytical methods, rather I use observational and logical techniques like yourself.
    I can't be of much help here, but I can lend a hand.

    How about if we tried dividing through by the highest power of x?

    The denominator will tend to 0 from the negative end, thus negative infinite?
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