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?

Yes, you are right. You want to use the identity $$a^2-b^2=(a+b)(a-b)$$ In your case $a:=\sqrt{x^2-6x+13}$ and $b=2$. The expansion is $$\left(\sqrt{x^2-6x+13}-2\right)\cdot\left(\sqrt{x^2-6x+13}+2\right)=x^2-6x+13-4=x^2-6x+9$$ Now, this simplifies significantly with the term in the numerator (but do not forget to multiply the numerator with the conjugate as well).
