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Thread: trig limit

  1. #1
    Senior Member polymerase's Avatar
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    trig limit

    Find the value of this limit using the fastest non-l'Hopital method...

    $\displaystyle \displaystyle\lim_{x\to{0}}\frac{x^2sec\;x}{sec\;x-1}$

    Thanks!
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by polymerase View Post
    $\displaystyle \displaystyle\lim_{x\to{0}}\frac{x^2sec\;x}{sec\;x-1}$

    Thanks!
    You may assume that $\displaystyle \lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12.$

    Now, multiply top & bottom by $\displaystyle \cos x.$
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  3. #3
    Super Member PaulRS's Avatar
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    $\displaystyle \lim_{x\rightarrow{0}}{\frac{x^2\cdot{\sec(x)}}{\s ec(x)-1}}=\lim_{x\rightarrow{0}}{\frac{x^2\cdot{\frac{1} {\cos(x)}}}{\frac{1}{\cos(x)}-1}}$

    THen: $\displaystyle \lim_{x\rightarrow{0}}{\frac{x^2\cdot{\sec(x)}}{\s ec(x)-1}}=\lim_{x\rightarrow{0}}{\frac{x^2}{1-\cos(x)}}$

    Now we multiply numerator and denominator by $\displaystyle 1+\cos(x)$

    We get: $\displaystyle \lim_{x\rightarrow{0}}{\frac{x^2\cdot{\sec(x)}}{\s ec(x)-1}}=\lim_{x\rightarrow{0}}{\frac{x^2(1+\cos(x))}{1-\cos^2(x)}}$

    But $\displaystyle \sin^2(x)=1-\cos^2(x)$ and $\displaystyle (1+\cos(x))\rightarrow 2$ so: $\displaystyle \lim_{x\rightarrow{0}}{\frac{x^2\cdot{\sec(x)}}{\s ec(x)-1}}=2\lim_{x\rightarrow{0}}{\frac{x^2}{\sin^2(x)}} =2$

    PS- Krizalid is faster than me
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  4. #4
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    Just note that:$\displaystyle
    \frac{{x^2 \sec (x)}}{{\sec (x) - 1}} = \frac{{x^2 }}{{1 - \cos (x)}} = \frac{{x^2 \left( {1 + \cos (x)} \right)}}{{\sin ^2 (x)}}$.
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