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I can't find what is wrong when using L'Hospital's rule on this limit. Derivative in denominator is $$\frac{-5x^3+4x^2-\sqrt{4-x^2}(6x-4)+12x-8}{2\sqrt{(1-x)(4-x^2)}}$$ and in numerator is $\frac{2\sqrt{2(4-x^2)(1+x)}(2-\sqrt{4-x^2})-4x\sqrt{2(1-x^2)(1-x)}-\sqrt{2+x}(x^2-2x(2-x)-1)-\sqrt{(2+x)(1-x^2)}}{\sqrt{2(4-x^2)(1-x^2)}}$

This gives undefined statement $\frac{0}{0}$. I tried applying L'Hospital's rule again, but that again gives $\frac{0}{0}$

Limit should be $L=1$

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using Bernoulli \begin{align} x \to 0 \hspace{10mm} & {\color{Red}{(1+ax)^n \approx 1+anx} } \\ & \sqrt{1-x^2} = (1-x^2)^{\frac{1}{2}} \approx 1-\frac{1}{2}x^2 \\ & \sqrt{4-x^2}=\sqrt{4(1-\frac{x^2}{4}})=2(1-\frac{x^2}{4})^{\frac{1}{2}} \approx 2(1-\frac{1}{2} \frac{x^2}{4})=2-\frac{x^2}{4} \end{align} so by putting them in limit : \begin{align} L &= \lim_{x \to 0} \frac{\sqrt{2(2-x)} (1-\sqrt{1-x^2})-4\sqrt{1-x}(2-\sqrt{4-x^2})}{x\sqrt{1-x}(2-\sqrt{4-x^2})} \\ &= \lim_{x \to 0} \frac{\sqrt{2(2-x)} (1-(1-\frac{1}{2}x^2))-4\sqrt{1-x}(2-(2-\frac{x^2}{4}))}{x\sqrt{1-x}(2-(2-\frac{x^2}{4}))} \\ &= \lim_{x \to 0} \frac{\sqrt{2(2-x)} (\frac{1}{2}x^2)-4\sqrt{1-x}(\frac{x^2}{4})}{x\sqrt{1-x}(\frac{x^2}{4})} \\ &= \lim_{x \to 0} \frac{x^2(\frac{1}{2}\sqrt{2(2-x)}-\sqrt{1-x})}{\frac{x^3}{4}\sqrt{1-x}} \\ &= \lim_{x \to 0} \frac{(\sqrt{\frac{2(2-x)}{4}}-\sqrt{1-x})}{\frac{x}{4}\sqrt{1-x}} = \lim_{x \to 0} \frac{(\sqrt{1-\frac{x}{2}}-\sqrt{1-x})}{\frac{x}{4}\sqrt{1-x}} \end{align} use bernoulli again \begin{align} L = \lim_{x \to 0} \frac{(1-\frac{x}{4})-(1-\frac{x}{2})}{\frac{x}{4}\sqrt{1-x}} = \lim_{x \to 0} \frac{\frac{x}{4}}{\frac{x}{4}\sqrt{1-x}} \end{align} simplify $\frac{x}{4}$ $$\lim_{x \to 0} \frac{1}{\sqrt{1-x}} = 1.$$

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