Solving Simple Limit Problem Without Substitution

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I tutor high school students in Calc and the other day I came across this problem.

Limit of (1-Sqrt(x-2))/(x-3) as x->3

I tried coaching the student on how to simplify the expression and in the end I just showed him this substitution.

Let u=Sqrt(x-2)

Then

(1-Sqrt(x-2))/(x-3) = (1-u)/(u^2-1)

And the Limit becomes

Limit of (1-u)/(u^2-1)=-1/(u+1) as u->1 which is -1/2

He looked at me like I had just done some black magic. I explained substitution to him and why it worked, showed him a couple of other simple examples, and confirmed the answer numerically (like they do in basic calc books when the limit concept is first presented). I still don't think he is 100% convinced because they had not covered this in his class yet which leads to my question.

Can the original problem be solved without substitution?
 
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Well, you might do it this way:
\frac{(1-\sqrt{x-2})}{x-3}=\frac{(1-\sqrt{x-2})}{x-3}*1=\frac{(1-\sqrt{x-2})}{x-3}*\frac{(1+\sqrt{x-2})}{(1+\sqrt{x-2})}=\frac{3-x}{(1+\sqrt{x-2})*(x-3)}=-\frac{1}{(1+\sqrt{x-2})}
And so on..
 
Multiplying by the conjugate would do the trick, but personally, I feel the easiest way is L'Hospital's rule .
 
Yipe. For some reason I though 1-x would be the numerator. Boy do I feel sheepish. :) Thanks for the help!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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