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Math Help - One sided limit question

  1. #1
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    One sided limit question

    <br />
\frac{12sin^3x}{12x^3}<br />
    <br />
\lim x \to 0^-<br />

    Heres what I have...
    can be rewritten as:
    <br />
\frac{12(sinx)^3}{12x^3}=\frac{(sinx)^3}{x^3}=\fra  c{sinx}{x}*\frac{sinx}{x}*\frac{sinx}{x}=1*1*1=1<br />

    Up to this point, am I correct? How does the fact that it wants the limit as x approaches zero from the left? I only know how to work with these one-sided limits in a peicewise function. What do I do from here & what do I need to learn to solve one sided limits that are NOT in a peicewise function?

    Thanks!
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  2. #2
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    Quote Originally Posted by Vamz View Post
    <br />
\frac{12sin^3x}{12x^3}<br />
    <br />
\lim x \to 0^-<br />

    Heres what I have...
    can be rewritten as:
    <br />
\frac{12(sinx)^3}{12x^3}=\frac{(sinx)^3}{x^3}=\fra  c{sinx}{x}*\frac{sinx}{x}*\frac{sinx}{x}=1*1*1=1<br />

    Up to this point, am I correct? How does the fact that it wants the limit as x approaches zero from the left? I only know how to work with these one-sided limits in a peicewise function. What do I do from here & what do I need to learn to solve one sided limits that are NOT in a peicewise function?

    Thanks!
    \displaystyle \lim_{x \to 0^-} \frac{12 \sin^3 (x)}{12 x^3} = \lim_{x \to 0^+} \frac{12 \sin^3 (x)}{12 x^3} = \lim_{x \to 0} \frac{12 \sin^3 (x)}{12 x^3} = 1.
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  3. #3
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    Quote Originally Posted by Vamz View Post
    <br />
\frac{12sin^3x}{12x^3}<br />

    <br />
\lim x \to 0^-<br />

    Heres what I have...
    can be rewritten as:

    <br />
\frac{12(sinx)^3}{12x^3}=\frac{(sinx)^3}{x^3}=\fra  c{sinx}{x}*\frac{sinx}{x}*\frac{sinx}{x}=1*1*1=1<br />

    Up to this point, am I correct? How does the fact that it wants the limit as x approaches zero from the left? I only know how to work with these one-sided limits in a peicewise function. What do I do from here & what do I need to learn to solve one sided limits that are NOT in a piecewise function?

    Thanks!
    x=-u

    x\rightarrow\ 0^-\Rightarrow\ u\rightarrow\ 0^+

    Sinx=Sin(-u)=-Sinu

    \displaystyle\frac{Sinx}{x}=\frac{-Sinu}{-u}=\frac{Sinu}{u}
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