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Math Help - Evaluate the given limit, Help

  1. #1
    Junior Member Cyberman86's Avatar
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    Evaluate the given limit, Help

    Evaluate the Given Limit, this doesn't look complicated but I'm Stuck anyways how can i finish this 2 problems?

    / means over the other function
    -> is an arrow as x approaches....

    Lim x^2 - 16 / over sqr. root of X - 2
    x->4


    So far I have:
    (X-4) (X+4) / sqr. root of 4 - 2 then i don't know if its right or how to finish it.

    ------------------------------------------------------------------------------

    Lim sqr. root of 3 - t minus sqr. root of 3 / over t
    t->0


    I tried to multiply both sides by sqr. root of 3 - t plus sqr. root of 3

    then i got
    sqr. root of 3 - t / over sqr. root of 3 - t plus sqr. root of 3

    I don't know how to finish or if i even started right.


    If you can, can you please show the steps.
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  2. #2
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Cyberman86 View Post
    Evaluate the Given Limit, this doesn't look complicated but I'm Stuck anyways how can i finish this 2 problems?

    / means over the other function
    -> is an arrow as x approaches....

    Lim x^2 - 16 / over sqr. root of X - 2
    x->4


    So far I have:
    (X-4) (X+4) / sqr. root of 4 - 2 then i don't know if its right or how to finish it.

    ------------------------------------------------------------------------------

    Lim sqr. root of 3 - t minus sqr. root of 3 / over t
    t->0


    I tried to multiply both sides by sqr. root of 3 - t plus sqr. root of 3

    then i got
    sqr. root of 3 - t / over sqr. root of 3 - t plus sqr. root of 3

    I don't know how to finish or if i even started right.


    If you can, can you please show the steps.
    use x^{2}-16=(x-4)(x+4)=(\sqrt{x}-2)(\sqrt{x}+2)(x+4) for the first one, I dont understand the second one... why dont you try using paranthesis?
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Cyberman86 View Post
    Evaluate the Given Limit, this doesn't look complicated but I'm Stuck anyways how can i finish this 2 problems?

    / means over the other function
    -> is an arrow as x approaches....

    [snip]

    Lim sqr. root of 3 - t minus sqr. root of 3 / over t
    t->0


    I tried to multiply both sides by sqr. root of 3 - t plus sqr. root of 3

    then i got
    sqr. root of 3 - t / over sqr. root of 3 - t plus sqr. root of 3

    I don't know how to finish or if i even started right.


    If you can, can you please show the steps.
    Is this what you mean? \lim_{t\to{0}}\frac{\sqrt{3-t}-\sqrt{3}}{t}?

    --Chris
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  4. #4
    Junior Member Cyberman86's Avatar
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    yes that's the exact problem but i don't how to write like that in here.
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  5. #5
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    Hello, Cyberman86!

    \lim_{x\to4} \frac{x^2 - 16}{\sqrt{x} - 2}

    Factor twice: . \lim_{x\to4} \frac{x^2-16}{\sqrt{x}-2} \;=\;\lim_{x\to4}\frac{(x-4)(x+4)}{\sqrt{x} - 2} \;=\; \lim_{x\to4}\frac{(\sqrt{x}-2)(\sqrt{x}+2)(x + 4)}{\sqrt{x}-2}

    Then: . \lim_{x\to4}\,(\sqrt{x}+2)(x+4) \;=\;(\sqrt{4}+2)(4+4) \;=\;32




    \lim_{t\to0}\frac{\sqrt{3-t} - \sqrt{3}}{t}

    Multiply top and bottom by (\sqrt{3-t} + \sqrt{3})

    \frac{\sqrt{3-t} - \sqrt{3}}{t}\cdot{\color{blue}\frac{\sqrt{3-t} + \sqrt{3}}{\sqrt{3-t} + \sqrt{3}}} \;=\;\frac{(3-t) - 3}{t(\sqrt{3-t} + \sqrt{3})} \;= . \frac{-{\color{red}\rlap{/}}t}{{\color{red}\rlap{/}}t(\sqrt{3-t} + \sqrt{3})}  \;=\;\frac{-1}{\sqrt{3-t} + \sqrt{3}}

    Then: . \lim_{x\to0}\frac{-1}{\sqrt{3-t} + \sqrt{3}} \;=\; \frac{-1}{\sqrt{3} + \sqrt{3}} \;=\;\frac{-1}{2\sqrt{3}}

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  6. #6
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,

    Another way for the second one.

    Recall the formula : \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}=f'(a)

    Here, f(x)=\sqrt{x} and a=3.
    f'(x)=\frac 1{2 \sqrt{x}}

    \lim_{t \to 0} \frac{\sqrt{3-t}-\sqrt{3}}{t}=\lim_{h \to 0} - \ \frac{\sqrt{3+h}-\sqrt{3}}{h}, by substituting h=-t

     =- \ \frac{f(3+h)-f(3)}{h}=-f'(3)=- \ \frac{1}{2 \sqrt{3}}
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