April 17th, 2018, 12:11 AM  #1 
Newbie Joined: Jul 2017 From: KOLKATA Posts: 29 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: 893 
When you have functions of x in both base and exponent, try taking the logarithm.

April 17th, 2018, 12:52 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,555 Thanks: 599 
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: 7,842 Thanks: 1068 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 
Newbie Joined: Jul 2017 From: KOLKATA Posts: 29 Thanks: 2 
Thanks . So it comes to 1


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calculus, limit 
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