1. Limited time only! Sign up for a free 30min personal 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!

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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper

    [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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook