How to find $$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x+x^2-2}$$
My Try :
$$x^2+x-2=(x-1)(x+2)$$
$$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{(x-1)(x+2)}\cdot \frac{(\sqrt{x}+\sqrt{x+3})+3}{(\sqrt{x}+\sqrt{x+3})+3}=\\ =\lim_{x \to 1} \frac{2x+2\sqrt{x(x+3)}-6}{(x-1)(x+2)((\sqrt{x}+\sqrt{x+3})+3)}=\\\lim_{x \to 1} \frac{2(x-3+\sqrt{x(x+3)})}{(x-1)(x+2)((\sqrt{x}+\sqrt{x+3})+3)} $$
Now what ?
On
First factor $x^2+x^2-2=(x-1)(x+2)$. $$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{(x-1)(x+2)}=$$
$$=\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x-1}\lim_{x\to 1}\frac{1}{x+2}=$$
$$=\left(\lim_{x\to 1}\left(\frac{\sqrt{x}-1}{x-1}+\frac{\sqrt{x+3}-2}{x-1}\right)\right)\cdot \frac{1}{3}$$
Hint: multiply the numerator and denominator of $\frac{\sqrt{x}-1}{x-1}$ by $\sqrt{x}+1$ and of $\frac{\sqrt{x+3}-2}{x-1}$ by $\sqrt{x+3}+2$.
Say $t=\sqrt{x}$, then
\begin{eqnarray} % \nonumber to remove numbering (before each equation) \lim_{t \to 1} \frac{t+\sqrt{t^2+3}-3}{t^4+t^2-2} &= &\lim_{t \to 1} \frac{t-1+\sqrt{t^2+3}-2}{(t^2+2)(t^2-1)} \\ &=& \lim_{t \to 1} \frac{1}{(t^2+2)(t+1)} + \frac{\sqrt{t^2+3}-2}{(t^2+2)(t^2-1)}\\ &=& {1\over 6}+\lim_{t \to 1} \frac{t^2-1}{(t^2+2)(t^2-1)(\sqrt{t^2+3}+2)}\\ &=& {1\over 6}+ {1\over 12}\\ &=& {1\over 4} \end{eqnarray}