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Find the imit of 2x + 1 - sqrt(4x^2 + 5) as x--> infinity

  1. Jan 29, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\lim_{x \rightarrow \infty}[/tex][tex] 2x+1-[/tex][tex]\sqrt{4x^2+5}[/tex]

    3. The attempt at a solution

    i am wondering if this method that i used is correct. i get the correct answer but i ahaven't see it in the text book or on the net. am i doing something that shouldn't be done?

    using the limit laws, i take out 1.

    [tex]\lim_{x \rightarrow \infty}[/tex] 1 + [tex]\lim_{x \rightarrow \infty [/tex][tex]2x-\sqrt{4x^2+5}[/tex]

    then i rationalize.
    [tex]\lim_{x \rightarrow \infty[/tex] 1 + [tex]\lim_{x \rightarrow \infty} \frac{4x^2-4x^2-5}{2x+\sqrt{4x^2+5}}[/tex]

    1 + [tex]\lim_{x \rightarrow \infty} [/tex][tex]\frac{-5}{2x+x\sqrt{4+\frac{5}{x^2}}}[/tex]

    1 + [tex]\lim_{x \rightarrow \infty} \frac{\frac{-5}{x}}{\frac{2x}{x}+\frac{x}{x}\sqrt{4x+\frac{5}{x^2}}}[/tex]

    1 + [tex]\frac{0}{4}[/tex]


    i get the right answer when i do it for [tex]\lim_{x \rightarrow \infty [/tex][tex]\sqrt{9x^4-3x^2+1} - 3x^2+5[/tex] and these the only two i tried. is this method just lucky for these two or can i keep using it?

    also. i seem to be getting the right limit of 1 when x approches postive infinity for [tex]\lim_{x \rightarrow \infty [/tex][tex]\sqrt{9x^2+6x-5} - 3x[/tex] but i get -1 when approches negative infinity and mathematica keeps telling me its +infinity. can someone do provide me with a soloution?
    Last edited: Jan 30, 2008
  2. jcsd
  3. Jan 29, 2008 #2
    I have no idea what your problem says, plz fix your latex!!!

    infinity = \infty

    lim as x approaches a = \lim_{x \rightarrow a}
  4. Jan 30, 2008 #3
  5. Jan 30, 2008 #4
    try pulling the x^2 out of the square root and then so how that affects the terms inside the square root as x goes to inf.
  6. Jan 30, 2008 #5
    i tried to solve it, but i keep on ending up with a zero at the bottom. does this mean that what i did is wrong or the limit is infinity whenever you end up with a zero in the denominator?
    THE INFINTY IS NEGATIVE. sorry i dont know to syntex of a negative.

    [tex]\lim_{x \rightarrow- \infty [/tex][tex]\sqrt{9x^2+6x-5} - 3x[/tex]

    [tex]\lim_{x \rightarrow- \infty [/tex][tex]\sqrt{9x^2+6x-5} - 3x[/tex] . [tex]\frac{\sqrt{9x^2+6x-5}+3x}{\sqrt{9x^2+6x-5}+3x}[/tex]

    [tex]\lim_{x \rightarrow- \infty [/tex] [tex]\frac{6x-5}{\sqrt{9x^2+6x-5}+3x}[/tex]

    [tex]\lim_{x \rightarrow- \infty [/tex] [tex]\frac{6x-5}{|x|\sqrt{9+\frac{6}{x}-\frac{6}{x^2}}+3x}[/tex]

    [tex]\lim_{x \rightarrow- \infty [/tex] [tex]\frac{6-\frac{5}{x}}{\frac{|x|}{x}\sqrt{9+\frac{6}{x}-\frac{6}{x^2}}+\frac{3x}{x}}[/tex]



    so does this mean the what i did is wrong, or does this indicate the limit is infinity like mathematica keep telling me and what i did is the correct steps?
    Last edited: Jan 30, 2008
  7. Jan 30, 2008 #6


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    Why is [tex]|x| / x = -1[/tex] if x goes to + infinity? What is |x| if x becomes very large?
    And my Mathematica doesn't tell me it's infinity.
  8. Jan 30, 2008 #7
    ^^^ oh i forgot to do it there but its going to negative infinity.

    i will fix it.

    i am getting the correct answer for all positive inf questions. just neg ones i am not getting.
  9. Jan 30, 2008 #8
    [tex]\lim_{x \rightarrow- \infty[/tex][tex]\sqrt{9x^2+6x-5} - 3x[/tex]

    can someone show me the proper soloution. i tried to find some explanation on the web but no help. i keep getting a zero in the denominator when i use this method.
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