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April 17th, 2018, 12:11 AM   #1
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calculus Limit

Find lim (sinx)^(2tanx)
x-->0
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April 17th, 2018, 03:08 AM   #2
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When you have functions of x in both base and exponent, try taking the logarithm.
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April 17th, 2018, 12:52 PM   #3
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For x near 0 $(sinx)^{2tanx}$ can be approximated by $x^{2x}$.
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April 17th, 2018, 10:39 PM   #4
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It's an indeterminate form so we'll take a look at using L'Hopital's rule:

$$\exp\left(\frac{\log(\sin(x))}{1/(2\tan(x))}\right)=\exp(0)=1$$

Can you fill in the missing steps?
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April 17th, 2018, 11:27 PM   #5
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Thanks . So it comes to 1
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