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Math Help - Evaluate limits:

  1. #1
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    Evaluate limits:

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  2. #2
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    Quote Originally Posted by zaurm View Post
    Use the small angle approximation for the first one. This says that for very small angles  \sin(kx) = kx , which leaves you with  \frac{4x}{7x} = \frac{4}{7}
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  3. #3
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    Hello, zaurm!

    a)\;\;\lim_{x\to0}\frac{\sin(4x)}{7x}
    This one uses the theorem: . \lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1

    Multiply by \tfrac{4}{4}\!:\quad \lim_{x\to0}\left(\frac{4}{4}\cdot\frac{\sin(4x)}{  7x}\right) \;=\;\lim_{x\to0}\left(\frac{4}{7}\cdot\frac{\sin(  4x)}{4x}\right) \;= . \frac{4}{7}\cdot\lim_{x\to0}\left(\frac{\sin(4x)}{  4x}\right) \;=\;\frac{4}{7}\cdot 1 \;=\;\frac{4}{7}




    b)\;\;\lim_{x\to\infty}\frac{2x^2-1}{7x^2+x-7}
    Divide numerator and denominator by x^2

    . . \lim_{x\to\infty}\frac{\frac{2x^2}{x^2} - \frac{1}{x^2}}{\frac{7x^2}{x^2} + \frac{x}{x^2} - \frac{7}{x^2}} \;=\;\lim_{x\to\infty}\frac{2 -\frac{1}{x^2}}{7 + \frac{1}{x} - \frac{7}{x^2}} \;= . \frac{2-0}{7+0-0} \;=\;\frac{2}{7}

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