<s>[INST] How to find $\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x+x^2-2}$? 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 ?
 [/INST] 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}
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