anonymous
  • anonymous
Find the limit of sin(4x)/sin(2x) as x approaches to 0-.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
So I got 4cos(4x)/2cos(2x)=2, is this right?
ybarrap
  • ybarrap
yes, but you can also use that sin(x) is about equal to x for small x
anonymous
  • anonymous
\[\frac{\sin(4x)}{4x}\times \frac{2x}{\sin(2x)}\times 2\]

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anonymous
  • anonymous
get \(1\times 1\times 2=2\) w/o l'hopital basically this is what @ybarrap said
anonymous
  • anonymous
So what's the limit of tanx/x^2 as x approaches to 0-? Isn't it sec^2 x/2x?
zepdrix
  • zepdrix
Lot of ways to approach this one :) You could also apply Sine Double Angle and change the top to 2sin(2x)cos(2x)
anonymous
  • anonymous
@Idealist that is a start, but you see now that the limit is undefined since the denominator goes to zero but the numerator goes to one
anonymous
  • anonymous
Or can I do (sinx/cosx)/x^2 and then take the derivative?
zepdrix
  • zepdrix
Ya I like your first approach.\[\Large \lim_{x\to0^-}\frac{\tan x}{x^2} \to \frac{0}{0}\]Taking the derivative gave you,\[\Large \lim_{x\to0^-}\frac{\sec^2x}{2x}\]
zepdrix
  • zepdrix
So what is the limit giving you now? :)

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