Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

What is $a$ and $b$?

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

Thanks

share|improve this question

2 Answers 2

up vote 0 down vote accepted

\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 $$2 \sqrt{b+1} = 4 \implies \sqrt{b+1}=2 \implies b = 3 \, \& \, a = 5$$

share|improve this answer

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?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.