<s>[INST] Studying $\lim_{x \rightarrow 0^+} \frac{xe^{-2x^2}-\sin(x)+\beta x^3}{x^{\alpha}\cos(x^2)}$ Study the following limit with $\alpha$ and $\beta$ parameters:
$$\lim_{x \rightarrow 0^+} \frac{xe^{-2x^2}-\sin(x)+\beta x^3}{x^{\alpha}\cos(x^2)}$$
My attempt:
$$\frac{xe^{-2x^2}-\sin(x)+\beta x^3}{x^{\alpha}\cos(x^2)} \;\sim_{0}\; \frac{xe^{-2x^2}-\sin(x)+\beta x^3}{x^{\alpha}}$$
Now I use Hospital three times and I have:
$$\frac{(-12+96x^2-64x^4)e^{-2x^2}+\cos(x)+6 \beta}{\alpha (\alpha-1)(\alpha-2)x^{\alpha-3}}\;\sim_0\; \frac{-11 +6 \beta}{\alpha (\alpha-1)(\alpha-2)x^{\alpha-3}}$$
Now if $\beta = \frac{11}{6}$ and $\alpha \gt 3$, I have $0 \cdot \infty$ indeterminate form. What can we say about the case $\beta = \frac{11}{6}$ and $\alpha \gt 3$?
 [/INST] $$xe^{-2x^2}-\sin x+\beta x^3$$$$=x\left(1-2x^2+\frac{(-2x^2)^2}2\right)-\left(x-\frac{x^3}6+\frac{x^5}{120}\right)+\beta x^3+o(x^5)$$$$\sim_{(x\to0)}\begin{cases}\left(\beta-\frac{11}6\right)x^3&\text{if }\beta\ne\frac{11}6\\\frac{239}{120}x^5&\text{if }\beta=\frac{11}6.\end{cases}$$
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