i'm having hard time with limx→∞x(1+ex2)tan⁡x−2sin2xx(tan⁡x−x)i've tried to simplify it using cos⁡2x identity (cos⁡2x=1−2sin2x→cos⁡2x−1=−2sin2x) ,or tried to break it into 2 different fractions, but i couldn't get it to a point where i could use mclaurin in order to solve the given limit.

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i&#039;m having hard time with limx→∞x(1+ex2)tan⁡x−2sin2xx(tan⁡x−x)i&#039;ve tried to simplify it using cos⁡2x identity (cos⁡2x=1−2sin2x→cos⁡2x−1=−2sin2x) ,or tried to break it into 2 different fractions, but i couldn&#039;t get it to a point where i could use mclaurin in order to solve the given limit.

annie996k88v · Accepted Answer

First the denominator:x(tan⁡x−x)≈x(x+x33−x)=x43We will probably need a fourth degree approximation at the numerator.Then the numerator:x(1+ex2)tan⁡x−2sin2x≈x(2+x2)x(1+x23)−2x2(1−x26)2=(23+1+2⋅26)x4+⋯Hence the limit7.Note that you can start by pulling out a factor sin⁡xx and evaluatex(1+ex2)sec⁡x−2sin⁡xtan⁡x−x≈x(2+x2)(1+x22)−2x(1−x26)x+x33−x=7x23+⋯x23

i'm having hard time with \(\displaystyle\lim_{{{x}\to\infty}}{\frac{{{x}{\left({1}+{e}^{{{x}^{{2}}}}\right)}{\tan{{x}}}-{2}{{\sin}^{{2}}{x}}}}{{{x}{\left({\tan{{x}}}-{x}\right)}}}}\) i've tried

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