Question

limits with maclaurin series, help someone

Answer 1

\[\lim_{x \rightarrow 0}\frac{ e ^{x ^{2}}-\ln(1+x ^{2})-1 }{ \cos(2x)+2x*\sin(x)-1 }\]

Answer 2

So I start to develop the denominator

\[\cos(2x) = 1-\frac{ (2x)^2 }{ 2! }+\frac{ (2x)^4 }{ 4! }+ordo(x ^{6})\]

\[2x*\sin(x)=2x(x-\frac{ x ^{3} }{ 3! }+\frac{ x ^{5} }{ 5! }+ordo(x ^{7}))\]

\[1-\frac{ (2x)^2 }{ 2! }+\frac{ (2x)^4 }{ 4! }+ordo(x ^{6}) +2x(x-\frac{ x ^{3} }{ 3! }+\frac{ x ^{5} }{ 5! }+ordo(x ^{7}))\]

Answer 3

is this correct?

Answer 4

@shamim

Answer 5

hmm its been so long - i've partly forgotten this stuff

Answer 6

yeah how about the numerator?

Answer 7

@ParthKohli how do you find how long you must expand?

Answer 8

that's a good question.

Answer 9

is there a trick or is just a good guess, and if its wrong just erase and startover?

Answer 10

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?

Answer 11

@ParthKohli @ganeshie8 @IrishBoy123