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Math Help - substitution of limits

  1. #1
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    substitution of limits

    how do you solve lim ( arcsin (3x) / tan (5x)) as x tends to 0 using substitution?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    how do you solve lim ( arcsin (3x) / tan (5x)) as x tends to 0 using substitution?
    First, a little work:

    \lim_{x\to0}\frac{\arcsin3x}{\tan5x}=\lim_{x\to0}\  frac{\arcsin3x\cos5x}{\sin5x}=\left(\lim_{x\to0}\f  rac{\arcsin3x}{\sin5x}\right)\cdot\left(\lim_{x\to  0}\cos5x\right)=\lim_{x\to0}\frac{\arcsin3x}{\sin5  x}

    Now let x=\frac{\sin y}{3}. It's clear that x\to0 implies y\to0.

    We have now:

    \lim_{x\to0}\frac{y}{\sin\left(\frac{5}{3}\sin y\right)}

    Can you take it from here?

    Hint: Multiply and divide by \frac{5}{3}\sin y and regroup the terms to something more amenable.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Or exactly the same as in your other post

    \lim_{x\to0}\frac{\arcsin(3x)}{\tan(5x)}=\lim_{x\t  o0}\frac{\tfrac{\arcsin(2x)-\arcsin(2\cdot0)}{x}}{\tfrac{\tan(5x)-\tan(5\cdot0)}{x}}=\frac{\bigg[\arcsin(2x)\bigg]'\bigg|_{x=0}}{\bigg[\tan(5x)\bigg]'\bigg|_{x=0}}

    You really need to try to to take some of the methods posted in your other quetsions and apply them, instead of mass posting every limit you see.
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