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Math Help - How do you evaluate limits at infinity?

  1. #1
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    How do you evaluate limits at infinity?

    I have a couple of questions asking me to evaluate limits at infinity, but I am a bit confused.

    These are the questions given in the example:

    Question: Evaluate limx->infinity (1) / (x+1)

    Question: Evaluate limx->infinity (x^3 + x^2 - 3) / (4x^3)

    So I'm not sure what the process is to solve questions like these so that I can use the same process in my homework questions.

    Any help is appreciated.
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  2. #2
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    Usually, one has to know of a few properties of division by infinity, infinity as an exponent etc. You can easily find a comprehensive list of such properties either online or in any calculus textbook. For your particular cases, "force factorization" will work.

    a) limx ->infinity 1/(x+1)=x(1/x)/x(1+1/x)=0 Here I factored both the denominator and the numerator by x; 1/x as x approaches inifnity is 0, so 0/(1+0)=0
    b)limx ->inifnity (x^3+x^2-3)/(4x^3)=x^3(1+(1/x)-(3/x^2))/x^3(4)=(1+0-0)/4=1/4
    Last edited by FlacidCelery; September 20th 2010 at 06:37 AM.
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  3. #3
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    Quote Originally Posted by iluvmathbutitshard View Post
    I have a couple of questions asking me to evaluate limits at infinity, but I am a bit confused.

    These are the questions given in the example:

    Question: Evaluate limx->infinity (1) / (x+1)

    Question: Evaluate limx->infinity (x^3 + x^2 - 3) / (4x^3)

    So I'm not sure what the process is to solve questions like these so that I can use the same process in my homework questions.

    Any help is appreciated.
    \displaystyle\lim_{x\rightarrow\infty}\frac{1}{x+1  }=\lim_{x\rightarrow\infty}\frac{1}{x\left(1+\frac  {1}{x}\right)}=\lim_{x\rightarrow\infty}\frac{1}{x  }\left(\frac{1}{1+\frac{1}{x}}\right)

    Now use the fact that \displaystyle\lim_{x\rightarrow\infty}\frac{1}{x}=  0


    \displaystyle\lim_{x\rightarrow\infty}\frac{x^3+x^  2-3}{4x^3}=\lim_{x\rightarrow\infty}\left(\frac{x^3}  {4x^3}+\frac{x^2}{4x^3}-\frac{3}{4x^3}\right)=\lim_{x\rightarrow\infty}\le  ft(\frac{1}{4}+\frac{1}{4x}-\frac{3}{4x^3}\right)
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  4. #4
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    That's more clear than text math, thanks Archie.
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