Question

Solve the limit as x approaches a of a-x/square root of x + square root of a

Answer 1

\[a-x/(\sqrt{x}+ \sqrt{a})\]

Answer 2

Direct substitution doesn't seem to be an issue here... Did you mean \(\sqrt x-\sqrt a\) in the denominator?

If you mean what you wrote, directly substituting will give your limit. No discontinuities to worry about. But that's boring. Another way you can approach this is factoring the difference of squares in the numerator:
\[\frac{a-x}{\sqrt x+\sqrt a}=-\frac{x-a}{\sqrt x+\sqrt a}=-\frac{\left(\sqrt x-\sqrt a\right)\left(\sqrt x+\sqrt a\right)}{\sqrt x+\sqrt a}=\sqrt a-\sqrt x\]
Now provided that \(a>0\), the limit is clearly 0.

Answer 3

Where did you get the a>0?

Answer 4

That's an assumption I'm making. If you have \(a=0\), then
\[\lim_{x\to a}\frac{a-x}{\sqrt x+\sqrt a}=\lim_{x\to 0}\frac{-x}{\sqrt x}=-\lim_{x\to0}\sqrt x\]
which does not exist as a two-sided limit.