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 April 17th, 2018, 12:11 AM #1 Member   Joined: Jul 2017 From: KOLKATA Posts: 30 Thanks: 2 calculus Limit Find lim (sinx)^(2tanx) x-->0
 April 17th, 2018, 03:08 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 When you have functions of x in both base and exponent, try taking the logarithm. Thanks from SDK
 April 17th, 2018, 12:52 PM #3 Global Moderator   Joined: May 2007 Posts: 6,613 Thanks: 617 For x near 0 $(sinx)^{2tanx}$ can be approximated by $x^{2x}$.
 April 17th, 2018, 10:39 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond 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?
 April 17th, 2018, 11:27 PM #5 Member   Joined: Jul 2017 From: KOLKATA Posts: 30 Thanks: 2 Thanks . So it comes to 1 Thanks from greg1313

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