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The question is compute $$ \lim_{x \to 0}\ \dfrac{x^2e^{x^2}+\dfrac{1}{2}\log(1+2x^2)-2x^2}{1+\arctan(x^6)-\cos(x^3)} $$ using Taylor series expansion around the origin, you should not use L'Hopital's rule. I tried substitute the Taylor series in the equation can cancel out a few terms but I still have no idea what to do.

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The necessary expansions for the denominator are \begin{align} \arctan t&=t-t^3+o(t^3) \\ \cos t&=1-\frac{t^2}{2!}+\frac{t^4}{4!}+o(t^4) \end{align} Therefore the denominator is $$ 1+x^6-x^{18}+o(x^{18})-1+\frac{x^6}{2!}-\frac{x^{12}}{4!}+o(x^{12})=\frac{3}{2}x^6+o(x^6) $$ Thus we know we have to stop expansions in the numerator at degree $6$: \begin{align} x^2e^{x^2}&=x^2\Bigl(1+x^2+\frac{x^4}{2}+o(x^4)\Bigr)=x^2+x^4+\frac{x^6}{2}+o(x^6) \\ \log(1+2x^2)&=2x^2-\frac{(2x^2)^2}{2}+\frac{(2x^2)^3}{3}+o((2x^2)^3)=2x^2-2x^4+\frac{8}{3}x^6+o(x^6) \end{align} Can you finish up?

The numerator is $$x^2+x^4+\frac{x^6}{2}+x^2-x^4+\frac{4}{3}x^6-2x^2+o(x^6)=\frac{11}{6}x^6+o(x^6)$$

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  • $\begingroup$ Thanks a lot. So the numerator is $\dfrac{11}{6}x^6$ and the denominator is $\dfrac{3}{2}x^6$ and therefore limit is $\dfrac{11}{9}$ I think it's a correct answer. Thanks. $\endgroup$ – Kyle Oct 16 '18 at 22:56
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To make life easier, start defining $y=x^2$ for numerator and $z=x^3$ for denominator.

So, the numerator becomes $$\text{num}= y\,e^{y}+\dfrac{1}{2}\log(1+2y)-2y$$ and the denominator $$\text{den}= 1+\arctan(z^2)-\cos(z)$$ Now, using Taylor for each piece $$\text{num}=y\left(1+y+\frac{y^2}{2}+\frac{y^3}{6}+O\left(y^4\right) \right) +\dfrac{1}{2}\left(2 y-2 y^2+\frac{8 y^3}{3}+O\left(y^4\right) \right)-2y=\frac{11 y^3}{6}+O\left(y^4\right)$$ $$\text{den}=1+\left(z^2+O\left(z^6\right) \right)-\left(1-\frac{z^2}{2}+\frac{z^4}{24}+O\left(z^6\right) \right)=\frac{3 z^2}{2}-\frac{z^4}{24}+O\left(z^6\right)=\frac{3 z^2}{2}+O\left(z^4\right)$$

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  • $\begingroup$ +1 for making life easier. $\endgroup$ – Paramanand Singh Oct 20 '18 at 7:09
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Three hints for you: 1) $\log (1+t) \sim t - \frac{t^2}{2} + O(t^3) $ as $ t \to 0$

2) $\cos t \sim 1-\frac{t^2}{2} +O(t^4)$ as $t \to 0$.

3) $e^t \sim 1 + t + \frac{t^2}{2} + O( t^3)$ as $t \to 0$

Can you handle from here?

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By Taylor's expansion to the $6^{th}$ order we obtain

$$\dfrac{x^2e^{x^2}+\dfrac{1}{2}\log(1+2x^2)-2x^2}{1+\arctan(x^6)-\cos(x^3)}=$$

$$=\dfrac{x^2\left(1+x^2+\frac12 x^4+o(x^4)\right)+\dfrac{1}{2}\left(2x^2-2x^4+\frac83 x^6+o(x^6)\right)-2x^2}{1+(x^6+o(x^6))-\left(1-\frac12x^6+o(x^6)\right)}=$$

$$=\dfrac{\frac{11}6 x^6+o(x^6)}{\frac32x^6+o(x^6)}=\dfrac{\frac{11}6 +o(1)}{\frac32+o(1)} \to\frac{11}9$$

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