karatechopper
  • karatechopper
Find the limit. (Special Trig functions) As the lim (sinx/2x^2-x) approaches 0.
Mathematics
jamiebookeater
  • jamiebookeater
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karatechopper
  • karatechopper
I broke the limit down and did this: sinx sinx ---- - ----- x 2x-1 I used the special trig function to find out that sinx/x=1. So now I am left with 1 - sinx/2x-1 And I'm stuck!
karatechopper
  • karatechopper
to make more sense- I am stuck on 1- (sinx/2x-1)
IrishBoy123
  • IrishBoy123
\[lim_{x\rightarrow 0} \frac{sin \ x}{2x^2 - x} = lim_{x\rightarrow 0} \frac{\frac{sin \ x}{x}}{2x - 1} \] plus \[lim_{x\rightarrow 0} \frac{sin}{ x} = 1\]

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ChillOut
  • ChillOut
Still having trouble?
karatechopper
  • karatechopper
yeah I still don't understand..
IrishBoy123
  • IrishBoy123
\(\large \frac{1}{2(0) - 1} = ???\)
ChillOut
  • ChillOut
You need to notice that 2x²-x = x(2x-1). That's what @IrishBoy123 did.
ChillOut
  • ChillOut
And that's the same as \(\frac{ \sin x }{ x }*\frac{ 1 }{ 2x-1 }\)
IrishBoy123
  • IrishBoy123
the crucial thing here is that \[lim_{x\rightarrow 0} \frac{sin \ x}{x} = 1\] so you set it up to have a \( \frac{sin \ x}{x} \) term but that \( \frac{sin \ x}{x} \) term is seminal
IrishBoy123
  • IrishBoy123
this the important bit @karatechopper ! https://gyazo.com/ad5c26d3cf8e2b41c20a8c7935b34a83 splitting it out so you have a \(\frac{sinx}{x}\) term
anonymous
  • anonymous
karatechopper
  • karatechopper
understood! thanks :)
karatechopper
  • karatechopper
BUT. How can you just take that x to the numerator when its being multiplied in the denominator?
ChillOut
  • ChillOut
When you have something like \(\LARGE\frac{\frac{a}{b}}{c}\) you can rewrite that as \(\LARGE\frac{a}{b*c}\).
ChillOut
  • ChillOut

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