<s>[INST] Show that $\dfrac{x^2-6x+9}{\sqrt{x^2-6x+13}-2}=\sqrt{(x-3)^2+4}+2$ and hence find the limit as x tends to 3. Show that $\dfrac{x^2-6x+9}{\sqrt{x^2-6x+13}-2}=\sqrt{(x-3)^2+4}+2$ and hence find the limit as x tends to 3.
How do I start with this question? Rationalising the denominator? But how would I expand it?
 [/INST] $$\frac{x^2-6x+9}{\sqrt{x^2-6x+13}-2}=\frac{(x-3)^2}{\sqrt{(x-3)^2+4}-2}$$ $$\frac{(x-3)^2}{\sqrt{(x-3)^2+4}-2}*\frac{\sqrt{(x-3)^2+4}+2}{\sqrt{(x-3)^2+4}+2}$$ $$\frac{(x-3)^2}{\sqrt{(x-3)^2+4}-2}*\frac{\sqrt{(x-3)^2+4}+2}{\sqrt{(x-3)^2+4}+2}=\frac{(x-3)^2\sqrt{(x-3)^2+4}+2(x-3)^2}{(x-3)^2}$$ $$\frac{(x-3)^2\sqrt{(x-3)^2+4}+2(x-3)^2}{(x-3)^2}=\sqrt{(x-3)^2+4}+2$$
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