# Simple limit problem with squares

I'm doing a refreshment course in math but I'm stuck with some problem. Although this problem doesn't look hard I don't know what I'm doing wrong.

$$\lim_{x \to 4} \frac{\sqrt{x-3}-1}{2\sqrt{2}-\sqrt{x^2-3x+4}}$$

I have tried to take the conjuct of both squares but I still got the indeterminate form of $\frac{0}{0}$.

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Note that $$\sqrt{x-3}-1=\frac{(\sqrt{x-3}-1)(\sqrt{x-3}+1)}{\sqrt{x-3}+1}=\frac{\color{red}{x-4}}{\sqrt{x-3}+1}$$ and that $$2\sqrt 2-\sqrt{x^2-3x+4}=\frac{(2\sqrt 2-\sqrt{x^2-3x+4})(2\sqrt 2+\sqrt{x^2-3x+4})}{2\sqrt 2+\sqrt{x^2-3x+4}}$$ $$=\frac{-\color{red}{(x-4)}(x+1)}{2\sqrt 2+\sqrt{x^2-3x+4}}.$$ Hence, we have $$\frac{\sqrt{x-3}-1}{2\sqrt 2-\sqrt{x^2-3x+4}}=\left(\frac{\color{red}{x-4}}{\sqrt{x-3}+1}\right)\div\left(\frac{-\color{red}{(x-4)}(x+1)}{2\sqrt 2+\sqrt{x^2-3x+4}}\right)$$ $$=-\frac{2\sqrt 2+\sqrt{x^2-3x+4}}{(x+1)(\sqrt{x-3}+1)}\to -\frac{2\sqrt 2+2\sqrt 2}{5\cdot 2}=-\frac{2\sqrt 2}{5}\ \ (x\to 4).$$
Multiply numerator and denominator by $\displaystyle (2\sqrt{2}+\sqrt{x^2-3x+4})(\sqrt{x-3}+1)$:
$$\lim_{x \to 4} \frac{\sqrt{x-3}-1}{2\sqrt{2}-\sqrt{x^2-3x+4}}=\lim_{x \to 4} \frac{ (2\sqrt{2}+\sqrt{x^2-3x+4})(\sqrt{x-3}+1)(\sqrt{x-3}-1)}{(2\sqrt{2}-\sqrt{x^2-3x+4}) (2\sqrt{2}+\sqrt{x^2-3x+4})(\sqrt{x-3}+1)}= \\ = \lim_{x \to 4}\frac{(2\sqrt{2}+\sqrt{x^2-3x+4})(x-3-1)}{-(x^2-3x-4)(\sqrt{x-3}-1)}=\lim_{x \to 4}\frac{(x-4)(2\sqrt{2}+\sqrt{x^2-3x+4})}{-(x+1)(x-4)(\sqrt{x-3}+1)}=\\=\lim_{x \to 4}\frac{(2\sqrt{2}+\sqrt{x^2-3x+4})}{-(x+1)(\sqrt{x-3}+1)}=\frac{-(2\sqrt{2}+2\sqrt{2})}{5(\sqrt{2}+1)}$$