anonymous
  • anonymous
Determine the limit (if it exists). lim sin(x)/5x as x-->0
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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Psymon
  • Psymon
Well, there is an identity: \[\lim_{x \rightarrow 0}\frac{ sinx }{ x }=1\]As long as the angle of sin and the variable in the denominator match, the whole term is 1. So you have sin(x) and a 5x in bottom. So you need to go and turn that 5x into just a regular x in a mathematically legitimate way. Once you can do that you can find your answer. Let meknow if youre unsure how.
anonymous
  • anonymous
I am unsure how yo do that @Psymon
Psymon
  • Psymon
Well, the usual method is to multiply top and bottom by the same thing. So in this case we would do this: |dw:1378862012186:dw| Kinda see what I did?

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anonymous
  • anonymous
Ohhhhh that makes so much sense! thank you so so much! :)
Psymon
  • Psymon
Yep, np ^_^
anonymous
  • anonymous
Also do you know what to do if the problem is lim sin^2(x)/x as x--> 0 @Psymon
Psymon
  • Psymon
Well. I believe we need to change that sin^2(x) into something with cosines. Might you have an idea how to do that?
anonymous
  • anonymous
1- cos^2(x)?
Psymon
  • Psymon
Bingo. Now you can factor that like a difference of squares :3
anonymous
  • anonymous
I completely forget how to factor that
Psymon
  • Psymon
Well, as long as you have anything minus anything, you can do this: \[(a-b) = (\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})\]
anonymous
  • anonymous
so (x-cos)(x+cos)?
Psymon
  • Psymon
Well, (1-cosx)(1+cosx)
Psymon
  • Psymon
So that means we have this: \[\frac{ (1-cosx)(1+cosx) }{ x } \]Now recall that we have this identity: \[\lim_{x \rightarrow 0}\frac{ 1-cosx }{ x }=0\]Now different books show something slightly different. Also keep in mind that: \[\lim_{x \rightarrow 0}\frac{ cosx-1 }{ x }=0\]Both are 0, its just different books or different professors tend to show one or the other. So again, we already have that top version in our equation: |dw:1378863833609:dw|
anonymous
  • anonymous
THANK YOU SO MUCH!
Psymon
  • Psymon
Yep, np ^_^

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