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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 $\alpha \gt 3$?

Blue
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asv
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1 Answers1

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$$xe^{-2x^2}-\sin x+\beta x^3$$$$=x\left(1-2x^2+\frac{(-2x^2)^2}2\right)-\left(x-\frac{x^3}6+\frac{x^5}{120}\right)+\beta x^3+o(x^5)$$$$\sim_{(x\to0)}\begin{cases}\left(\beta-\frac{11}6\right)x^3&\text{if }\beta\ne\frac{11}6\\\frac{239}{120}x^5&\text{if }\beta=\frac{11}6.\end{cases}$$

Anne Bauval
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