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Homework Help: Evaluating Limits Algebraically

  1. Jan 31, 2008 #1
    I've never simplified an expression with a subtraction of radicals in the numerator. I've done a bit of reading and I'm guessing the method to use would be conjugation, but I'm not sure how that would let me solve for x. Thanks for any help.

    [tex]\lim_{x\rightarrow 0} \frac{(\sqrt{x+5} - \sqrt{5})}x[/tex]

    I'm also having trouble with these two limit estimations. I've worked with [tex]\Delta[/tex] before but never in a simplification.

    \lim_{x\rightarrow 0} \frac{2(x + \Delta x)-2x}{\Delta x}

    \lim_{x\rightarrow 0} \frac{(x + \Delta x^2) - 2(x + \Delta x) + 1 - (x^2 - 2x + 1)}{\Delta x}
    Last edited: Jan 31, 2008
  2. jcsd
  3. Jan 31, 2008 #2
    Multiply both numerator and denominator by ...

    [tex](\sqrt{x+5}+\sqrt 5)[/tex]
  4. Jan 31, 2008 #3


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    As you suggested and rocophysics confirmed, multiply both numerator and denominator by the conjugate.

    Go ahead and do the algebra! This is
    [tex]\lim_{x\rightarrow 0}\frac{2x+ 2\Delta x- 2x}{\Delta x}= \lim_{x\rightarrow 0}\frac{2\Delta x}{\Delta x}[/itex]
    What is the limit of that?

    First, you have that typed wrong: it should be [itex](x+ \Delta x)^2[/itex], not [itex](x+ \Delta x^2)[/itex]!
    Again, just do the algebra:
    [tex]\lim_{x\rightarrow 0}\frac{x^2+ 2x\Delta x+ \Delta x^2- 2x- \Delta x+ 1- x^2+ 2x- 1}{\Delta x}[/tex]
    [tex]= \lim_{x\rightarrow 0}\frac{2x\Delta x+ \Delta x^2k}{\Delta x}[/tex]
    What is the limit of that?
  5. Jan 31, 2008 #4
    Thanks very much for your help Ivy.

    http://calcchat.tdlc.com/solutionart/calc8e/01/c/se01c01053.gif [Broken]

    I can get up to step 4, but how do I go from 4 -> 5? What rule/formula is being used?
    Last edited by a moderator: May 3, 2017
  6. Jan 31, 2008 #5
    multiply by sqrt(5)/sqrt(5)
  7. Jan 31, 2008 #6
    Wouldn't that leave me with [tex]\frac{\sqrt{5}}{2}[/tex]? Could you be a bit more specific?
  8. Jan 31, 2008 #7


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    [tex]2\sqrt{5} \times \sqrt{5}=2 (\sqrt{5})^2=2 (5) = 10[/tex]
  9. Jan 31, 2008 #8
    Gotchya, I was trying to cancel it out rather than square it. Thanks for the help.
  10. Feb 3, 2009 #9
    To expand on this topic...
    Ive not seen many examples of evaluating limits algebraically where there are indexes in the root...for example

    \lim_{x\rightarrow 64} \frac{\sqrt{x} - 8}{\sqrt[3]{x}-4}

    ....so...are there any good techniques or method to solve this, since you can't use the conjugate?

    thanks in advance...
  11. Feb 3, 2009 #10
    [tex]\lim_{x\rightarrow 64} \frac{\sqrt{x} - 8}{\sqrt[3]{x}-4} = \lim_{x\rightarrow 64} \frac{(x^{1/6} - 2)(x^{1/3} + 2x^{1/6} + 4)}{(x^{1/6} - 2)(x^{1/6} + 2)} = \lim_{x\rightarrow 64} \frac{x^{1/3} + 2x^{1/6} +4}{x^{1/6} +2}[/tex].

    Now you can evaluate the limit by substituting 64 for x.
  12. Feb 4, 2009 #11
    Thanks for the reply...thats much faster it seems than the way I came up with...even though the principles are the same:

    I take the LCM of the exponents
    and apply a substitution variable

    a^6 = 1

    which leads to...


    \lim_{a\rightarrow 1} \frac{\sqrt{a^6} - 8}{\sqrt[3]{a^6}-4}


    Then it kinda becomes a binomial theorem expansion problem:


    \lim_{a\rightarrow 1} \frac{a^3 - 8}{a^2 - 4}


    this seems drawn out compared to your way... sorry for the elementary question but how did you decompose the radicals and get the equation

    \lim_{x\rightarrow 64} \frac{(x^{1/6} - 2)(x^{1/3} + 2x^{1/6} + 4)}{(x^{1/6} - 2)(x^{1/6} + 2)}

    to simplify to substitute 64 for x?

    Again, thanks for the reply!
  13. Feb 4, 2009 #12
    I used a difference of cubes:

    [tex] a^3 - b^3 = (a - b)(a^2 + ab + b^2)[/tex]
  14. Feb 17, 2009 #13
    I showed your method for solving that problem to my instructor...she responded by giving me a problem that I am finding rather difficult to implement your method.
    Do you think you could help?
    Same type of problem, just bigger numbers and no special binomials I could find:


    \lim_{x\rightarrow 4095} \frac{\sqrt{x+1}- 4\sqrt[3]{x+1}}{\sqrt[4]{x+1}-8}


    Thanks again.
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