Evaluating $\lim_{x\to1}{\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}}$ without L'Hospital's Theorem I've been trying to evaluate$$\lim_{x\to1}{\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}}$$ 
I tried: 
(a) Rationizing the numerator -> Error 
(b) Rationizing the denominator -> Error 
(c) Factoring out $x$ -> Error 

Finally, I used L'Hospital's Rule and got the answer $3/2$, but that is not what I am supposed to do, for not a single word about this Theorem was mentioned during my lectures. 
Is there any other way to solve this limit without this Theorem?

$$\lim_\limits{x\to1} \frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}$$
$$=\lim_\limits{x\to1} \left[\frac{\frac{\sqrt{x^2+3}-2}{x^2-1}}{\frac{\sqrt{x^2+8}-3}{x^2-1}}\right]$$
$$=\lim_\limits{x\to1} \left[\frac{\frac{\sqrt{x^2+3}-2}{x^2+3-2}}{\frac{\sqrt{x^2+8}-3}{x^2+8-9}}\right]$$
$$=\frac{\lim_\limits{x^2+3\to4} \frac{\sqrt{x^2+3}-\sqrt4}{(x^2+3)-4}}{\lim_\limits{x^2+8\to9} \frac{\sqrt{x^2+8}-\sqrt9}{(x^2+8)-9}}$$
$$=\frac{\frac{1}{2}4^{\frac{1}{2}-1}}{\frac{1}{2}9^{\frac{1}{2}-1}}$$
$$=\frac{\frac{1}{4}}{\frac{1}{6}}$$
$$=\frac{3}{2}$$
