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$$\lim_{x \rightarrow 1} \frac{\sqrt{a-x} - \sqrt{b+x}}{x-1} = -\frac 1 2$$

What is $a$ and $b$?

I just need a general direction to solve these type of questions.


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\begin{align} f(x)&=\dfrac{\sqrt{a-x} - \sqrt{b+x}}{x-1}\\ & = \dfrac{(a-x)-(b+x)}{x-1} \cdot \dfrac1{\sqrt{a-x} + \sqrt{b+x}}\\ & = \dfrac{(a-b) - 2x}{x-1} \cdot \dfrac1{\sqrt{a-x} + \sqrt{b+x}}\\ & = -2 \cdot \left(\dfrac{x - \dfrac{a-b}2}{x-1} \right) \cdot \left(\dfrac1{\sqrt{a-x} + \sqrt{b+x}}\right) \end{align} For $\lim_{x \to 1}$ to exist, we need $\dfrac{a-b}2=1$ i.e. $a=b+2$. In which case, we get that $$f(x) = -\dfrac2{\sqrt{a-x} + \sqrt{b+x}} \,\,\,\,\,\, \forall x \neq 1$$ Hence, $$\lim_{x \to 1} f(x) = - \dfrac2{\sqrt{a-1} + \sqrt{b+1}} = -\dfrac12$$ This gives us $$\sqrt{a-1} + \sqrt{b+1} = 4$$ Since $a=b+2$, we get that $$2 \sqrt{b+1} = 4 \implies \sqrt{b+1}=2 \implies b = 3 \, \& \, a = 5$$

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To begin with, it must be that

$$\sqrt{a-x}-\sqrt{b+x}\xrightarrow [x\to 1]{}0$$

otherwise the limit cannot be finite, so assuming this and applying L'Hospital:

$$-\frac{1}{2}=\lim_{x\to 1}\frac{\sqrt{a-x}-\sqrt{b+x}}{x-1}=\lim_{x\to 1}\left(\frac{-1}{\sqrt{a-x}}-\frac{1}{\sqrt{b+x}}\right)=$$

$$=-\lim_{x\to 1}\frac{\sqrt{b+x}+\sqrt{a-x}}{\sqrt{a-x}\sqrt{b+x}}$$

Can you take it from here?

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