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Thread: Check my limits Solution

  1. #1
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    Check my limits Solution

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

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

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

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

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

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


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

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

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

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

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

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


    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

    $\displaystyle x^3-1 = (x-1)(x^2+x+1)$

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

    ... 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

    $\displaystyle x^3-1 = (x-1)(x^2+x+1)$

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

    ... 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 $\displaystyle x = t^3$ so that you have $\displaystyle \lim_{t \rightarrow 1} \frac{t - 1}{t^3 - 1}$. 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 $\displaystyle \sqrt[3]x$ as $\displaystyle x^{\frac13}$.

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

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

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

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

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

    $\displaystyle =\frac13$

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

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

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

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

    Grandad
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