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Math Help - Check my limits Solution

  1. #1
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    Check my limits Solution

    1)  lim (x\rightarrow 1) \frac{\sqrt{x}-1}{x-1}
     \frac{1}{\sqrt{x}+1} =\frac{1}{2}

    2) lim(x\rightarrow -1) \frac{\sqrt{h+5}-2}{h+1}
     ]\frac{1}{\sqrt{h+5}+2}=\frac{1}{4}

    3)  lim(x\rightarrow 1)\frac{^3\sqrt{x}-1}{x-1}

     \frac{1}{^3\sqrt{x}+1}=\frac{1}{2}

    4)  lim(x\rightarrow 0) \frac{^3\sqrt{h+1}-1}{h}

     \frac{1}{^3\sqrt{h+1}+1}=\frac{1}{2}


    If there are any mistake please correct them?

    Thank you..
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  2. #2
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    Hello mj.alawami
    Quote Originally Posted by mj.alawami View Post
    1)  lim (x\rightarrow 1) \frac{\sqrt{x}-1}{x-1}
     \frac{1}{\sqrt{x}+1} =\frac{1}{2}

    2) lim(x\rightarrow -1) \frac{\sqrt{h+5}-2}{h+1}
     ]\frac{1}{\sqrt{h+5}+2}=\frac{1}{4}

    3)  lim(x\rightarrow 1)\frac{^3\sqrt{x}-1}{x-1}

     \frac{1}{^3\sqrt{x}+1}=\frac{1}{2}

    4)  lim(x\rightarrow 0) \frac{^3\sqrt{h+1}-1}{h}

     \frac{1}{^3\sqrt{h+1}+1}=\frac{1}{2}


    If there are any mistake please correct them?

    Thank you..
    The first two are fine, but there's a problem with #3 and #4. In these you need to use

    x^3-1 = (x-1)(x^2+x+1)

    So #3 simplifies to \frac{1}{x^{\frac23}+x^{\frac13}+1}

    ... and #4 is similar.

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello mj.alawamiThe first two are fine, but there's a problem with #3 and #4. In these you need to use

    x^3-1 = (x-1)(x^2+x+1)

    So #3 simplifies to \frac{1}{x^{\frac23}+x^{\frac13}+1}

    ... and #4 is similar.

    Grandad
    Can you please show me the complete steps of rationalizing the numerator of #3 ?
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  4. #4
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    Quote Originally Posted by mj.alawami View Post
    Can you please show me the complete steps of rationalizing the numerator of #3 ?
    My advice for #3 is to make the substitution x = t^3 so that you have \lim_{t \rightarrow 1} \frac{t - 1}{t^3 - 1}. Now factorise the denominator, cancel the factor common to numerator and denominator and then take the limit.

    I advise #4 be done in a similar way.
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  5. #5
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    Hello mj.alawami

    Use Mr F's substitution, or you can do it directly, by writing \sqrt[3]x as x^{\frac13}.

    Note first what happens when you multiply out the brackets from (x^{\frac13}-1)(x^{\frac23}+x^{\frac13}+1), using the laws of indices:

    (x^{\frac13}-1)\color{red}(x^{\frac23}+x^{\frac13}+1)\color{bla  ck}= x^{\frac13}\color{red}(x^{\frac23}+x^{\frac13}+1)\  color{black}-1\color{red}(x^{\frac23}+x^{\frac13}+1)\color{blac  k}=x + x^{\frac23}+x^{\frac13}-x^{\frac23}-x^{\frac13}-1=x-1

    So we can factorise the denominator: x-1 =(x^{\frac13}-1)(x^{\frac23}+x^{\frac13}+1)

    So \lim_{x\to 1}\frac{\sqrt[3]x-1}{x-1}=\lim_{x\to 1}\frac{x^{\frac13}-1}{(x^{\frac13}-1)(x^{\frac23}+x^{\frac13}+1)}

    =\lim_{x\to 1}\frac{1}{x^{\frac23}+x^{\frac13}+1}

    =\frac13

    Question 4 can be done in the same way. Start with

    \lim_{h\to0}\frac{\sqrt[3]{h+1}-1}{h}

    =\lim_{h\to0}\frac{(h+1)^{\frac13}-1}{(h+1)-1}

    ...which is basically the same as #3, with x replaced by (h+1).

    Grandad
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