anonymous
  • anonymous
Limit help! limit of [((sin^2)x)/x] as x approaches 0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\lim_{x\to0}\frac{\sin^2x}{x}=\lim_{x\to0}\sin x\cdot\lim_{x\to0}\frac{\sin x}{x}\]
anonymous
  • anonymous
but when you plug zero into the sin, how do you know which sin to use? there's two sins with zeros
anonymous
  • anonymous
Not sure I understand your question. \(\sin^2x=\sin x\sin x\), so you have \[\lim_{x\to0}\frac{\sin x\sin x}{x}\] Then, using a property of limits, you have \[\lim_{x\to0}\sin x\cdot\lim_{x\to0}\frac{\sin x}{x}\] You have two limits here. The first is 0, since \(\sin0=0\). The second is 1, since \(\lim_{x\to0}\dfrac{\sin x}{x}=1\), a known limit. So all-in-all, your answer is 0, since 0∙1=0.

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anonymous
  • anonymous
Would you need the unit circle to find the sin of zero or is that just something I should know?
anonymous
  • anonymous
Yeah, you should be familiar with \(\sin0=0\). Do you know about the second limit though? I would think that's the more important aspect of the problem.
anonymous
  • anonymous
..a second limit?
anonymous
  • anonymous
The sinx/x one, I mean. Have you learned that \[\lim_{x\to0}\frac{\sin x}{x}=1~~?\]
anonymous
  • anonymous
Yes, and I also learned that (1-cosx)/x = 1
anonymous
  • anonymous
whoops! that should be = 0
anonymous
  • anonymous
Right, well this problem is just making you apply that knowledge.
anonymous
  • anonymous
Could you help me with another limit problem?
anonymous
  • anonymous
Yeah
anonymous
  • anonymous
what is the limit of (cos x/cot x) as x approaches pi/2 ?
anonymous
  • anonymous
\[\frac{\cos x}{\cot x}=\frac{\cos x}{\frac{\cos x}{\sin x}}=\cos x\frac{\sin x}{\cos x}=\sin x\] So what's the limit \[\lim_{x\to\pi/2}\sin x~~?\]
anonymous
  • anonymous
wait, when you changed the cot x in the denominator, what did you change it to? I can't see it xD
anonymous
  • anonymous
\[\cot x=\frac{\cos x}{\sin x}\]
anonymous
  • anonymous
the sin of pi/2 is 1
anonymous
  • anonymous
Correct!
anonymous
  • anonymous
I have one last problem if you don't mind?

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