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Sizing integrals like fractions; name of paper
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LSpice
LSpice
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$\dfrac $\displaystyle\dfrac{a(x)}{x^{2}}\int_{0}^{x}b(\tau)d\tau \sim \dfrac{2}{1-\alpha}\dfrac{\sqrt{a(x)}}{x^{3/2}}, \ \text{when} \ x \to 0.$

Knowing that $b,a \in C^{0}((0,L]) \cap C^{1}((0,L))$, are positive and $b(x) = \dfrac{1}{\sqrt{xa(x)}}$. Assume that $0 < \alpha < 1$ and $$ \int_{0}^{x}b(\tau)d\tau \sim \dfrac{2}{1-\alpha}\dfrac{\sqrt{x}}{\sqrt{a(x)}}, \ \ \text{when} \ \ x \to 0 $$ Then $$\dfrac{a(x)}{x^{2}}\int_{0}^{x}b(\tau)d\tau \sim \dfrac{2}{1-\alpha}\dfrac{\sqrt{a(x)}}{x^{3/2}}, \ \ \text{when} \ \ x \to 0.$$

I'm trying to understand the above implication which is a passage from the proof of Lemma 5.12 of this article

https://link.springer.com/article/10.1007/s00028-006-0214-6 Martinez & Vancostenoble - Carleman estimates for one-dimensional degenerate heat equations.

It seems to be simple, I'm trying to apply L'hopital's rule. Basically what the author of the article did was multiply $\dfrac{a(x)}{x^{2}}$ by the limit of $\int_{0}^{x}b(\tau)d\tau$$\displaystyle\int_{0}^{x}b(\tau)d\tau$. Why is this possible?

$\dfrac{a(x)}{x^{2}}\int_{0}^{x}b(\tau)d\tau \sim \dfrac{2}{1-\alpha}\dfrac{\sqrt{a(x)}}{x^{3/2}}, \ \text{when} \ x \to 0.$

Knowing that $b,a \in C^{0}((0,L]) \cap C^{1}((0,L))$, are positive and $b(x) = \dfrac{1}{\sqrt{xa(x)}}$. Assume that $0 < \alpha < 1$ and $$ \int_{0}^{x}b(\tau)d\tau \sim \dfrac{2}{1-\alpha}\dfrac{\sqrt{x}}{\sqrt{a(x)}}, \ \ \text{when} \ \ x \to 0 $$ Then $$\dfrac{a(x)}{x^{2}}\int_{0}^{x}b(\tau)d\tau \sim \dfrac{2}{1-\alpha}\dfrac{\sqrt{a(x)}}{x^{3/2}}, \ \ \text{when} \ \ x \to 0.$$

I'm trying to understand the above implication which is a passage from the proof of Lemma 5.12 of this article

https://link.springer.com/article/10.1007/s00028-006-0214-6

It seems to be simple, I'm trying to apply L'hopital's rule. Basically what the author of the article did was multiply $\dfrac{a(x)}{x^{2}}$ by the limit of $\int_{0}^{x}b(\tau)d\tau$. Why is this possible?

$\displaystyle\dfrac{a(x)}{x^{2}}\int_{0}^{x}b(\tau)d\tau \sim \dfrac{2}{1-\alpha}\dfrac{\sqrt{a(x)}}{x^{3/2}}, \ \text{when} \ x \to 0.$

Knowing that $b,a \in C^{0}((0,L]) \cap C^{1}((0,L))$, are positive and $b(x) = \dfrac{1}{\sqrt{xa(x)}}$. Assume that $0 < \alpha < 1$ and $$ \int_{0}^{x}b(\tau)d\tau \sim \dfrac{2}{1-\alpha}\dfrac{\sqrt{x}}{\sqrt{a(x)}}, \ \ \text{when} \ \ x \to 0 $$ Then $$\dfrac{a(x)}{x^{2}}\int_{0}^{x}b(\tau)d\tau \sim \dfrac{2}{1-\alpha}\dfrac{\sqrt{a(x)}}{x^{3/2}}, \ \ \text{when} \ \ x \to 0.$$

I'm trying to understand the above implication which is a passage from the proof of Lemma 5.12 of Martinez & Vancostenoble - Carleman estimates for one-dimensional degenerate heat equations.

It seems to be simple, I'm trying to apply L'hopital's rule. Basically what the author of the article did was multiply $\dfrac{a(x)}{x^{2}}$ by the limit of $\displaystyle\int_{0}^{x}b(\tau)d\tau$. Why is this possible?

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user253963
user253963
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$\dfrac{a(x)}{x^{2}}\int_{0}^{x}b(\tau)d\tau \sim \dfrac{2}{1-\alpha}\dfrac{\sqrt{a(x)}}{x^{3/2}}, \ \text{when} \ x \to 0.$

Knowing that $b,a \in C^{0}((0,L]) \cap C^{1}((0,L))$, are positive and $b(x) = \dfrac{1}{\sqrt{xa(x)}}$. Assume that $0 < \alpha < 1$ and $$ \int_{0}^{x}b(\tau)d\tau \sim \dfrac{2}{1-\alpha}\dfrac{\sqrt{x}}{\sqrt{a(x)}}, \ \ \text{when} \ \ x \to 0 $$ Then $$\dfrac{a(x)}{x^{2}}\int_{0}^{x}b(\tau)d\tau \sim \dfrac{2}{1-\alpha}\dfrac{\sqrt{a(x)}}{x^{3/2}}, \ \ \text{when} \ \ x \to 0.$$

I'm trying to understand the above implication which is a passage from the proof of Lemma 5.12 of this article

https://link.springer.com/article/10.1007/s00028-006-0214-6

It seems to be simple, I'm trying to apply L'hopital's rule. Basically what the author of the article did was multiply $\dfrac{a(x)}{x^{2}}$ by the limit of $\int_{0}^{x}b(\tau)d\tau$. Why is this possible?