Csdrhrt

I need to prove

$$lim_{xto2}dfrac{sqrt{6−x}−2}{sqrt{3−x}−1}$$

I know the limit is 1/2, but I can't seem to manipulate the absolute value properly to get |x-2|*(something I can find an upper bound for), partially because simplifying then multiplying the top and bottom by the bottom conjugate gets |x-2| on the bottom

You do a great job by multiplying the conjugate of the bottom

Hint: Do that again for the top one and it will make the expression much easier to handle.

We have to show, $|frac{sqrt{6-x}-2}{sqrt{3-x}-1} - frac{1}{2}|< epsilon$ whenever $|x - 2|<delta(epsilon)$.

$|frac{sqrt{6-x}-2}{sqrt{3-x}-1} - frac{1}{2}|=|frac{sqrt{3-x}+1}{sqrt{6-x}+2} - frac{1}{2}|$ $[because xne 2]$ $= |frac{2sqrt{3-x} - sqrt{6-x}}{2(sqrt{6-x} + 2)}| = |frac{3(2-x)}{2(sqrt{6-x} + 2)(2sqrt{3-x} + sqrt{6-x})}|$

if $delta < 1$, then $|frac{3(2-x)}{2(sqrt{6-x} + 2)(2sqrt{3-x} + sqrt{6-x})}| < frac{3}{4}|x-2|$

Now choose, $delta = min{frac{1}{2},frac{4epsilon}{3}}$

Hint:

Notice that using the conjugates,

$$dfrac{sqrt{6−x}−2}{sqrt{3−x}−1}=frac{2-x}{2-x}dfrac{sqrt{3−x}+1}{sqrt{6−x}+2}$$ and the first fraction disappears.

The remaining one is continuous (the undeterminacy is gone) so that the limit is $f(2)=dfrac12$.

$$lim_{xto2}frac{sqrt{6-x}-2}{sqrt{3-x}-1}$$

$$lim_{xto2}frac{2-x}{2-x}cdotfrac{sqrt{3-x}+1}{sqrt{6-x+2}}$$

$$lim_{xto2}frac{sqrt{3-x}+1}{sqrt{6-x+2}}$$

$$=frac{2}{4}=frac{1}{2}$$