anonymous
  • anonymous
limits with maclaurin series, help someone
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\lim_{x \rightarrow 0}\frac{ e ^{x ^{2}}-\ln(1+x ^{2})-1 }{ \cos(2x)+2x*\sin(x)-1 }\]
anonymous
  • anonymous
So I start to develop the denominator \[\cos(2x) = 1-\frac{ (2x)^2 }{ 2! }+\frac{ (2x)^4 }{ 4! }+ordo(x ^{6})\] \[2x*\sin(x)=2x(x-\frac{ x ^{3} }{ 3! }+\frac{ x ^{5} }{ 5! }+ordo(x ^{7}))\] \[1-\frac{ (2x)^2 }{ 2! }+\frac{ (2x)^4 }{ 4! }+ordo(x ^{6}) +2x(x-\frac{ x ^{3} }{ 3! }+\frac{ x ^{5} }{ 5! }+ordo(x ^{7}))\]
anonymous
  • anonymous
is this correct?

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anonymous
  • anonymous
@shamim
welshfella
  • welshfella
hmm its been so long - i've partly forgotten this stuff
ParthKohli
  • ParthKohli
yeah how about the numerator?
anonymous
  • anonymous
@ParthKohli how do you find how long you must expand?
ParthKohli
  • ParthKohli
that's a good question.
anonymous
  • anonymous
is there a trick or is just a good guess, and if its wrong just erase and startover?
anonymous
  • anonymous
I changed to only two steps so the numerator is \[e ^{x ^{2}}=1+x ^{2}+\frac{ (x ^{2})^{2} }{ 2 }+ordo(x ^{3})\] \[\ln(1+x ^{2})=x ^{2}-\frac{ (x^{2})^{2} }{ 2 }+ordo(x ^{3})\] simplifying both numerator and determinator we have: \[\frac{ 1+x ^{4}+ordo(x ^{3}) }{ 1-\frac{ x ^{4} }{ 3 }+ordo(x ^{4}) }\] breaking out dominating term, when limes is -> 0 its the lowest exponent in numerator and determinator so its x^4 \[\frac{ x^4 }{ x^4 }*\frac{ \frac{ 1 }{ x^4 }+1+ordo(x^3) }{ \frac{ 1 }{ x^4 } -\frac{ x^4 }{ \frac{ 3 }{ x^4 } }+ordo(x^3) }\] and left we have -3... but mathematica says its 3, what have i done wrong?
anonymous
  • anonymous
@ParthKohli @ganeshie8 @IrishBoy123

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