<s>[INST] 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?
 [/INST] Notice, $$\lim_{x\to 1}\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}$$
$$=\lim_{x\to 1}\frac{(\sqrt{x^2+3}-2)(\sqrt{x^2+3}+2)}{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}\cdot \frac{(\sqrt{x^2+8}+3)}{(\sqrt{x^2+3}+2)}$$
$$=\lim_{x\to 1}\frac{x^2+3-4}{x^2+8-9}\cdot \frac{(\sqrt{x^2+8}+3)}{(\sqrt{x^2+3}+2)}$$
$$=\lim_{x\to 1}\frac{(x^2-1)}{(x^2-1)}\cdot \frac{(\sqrt{x^2+8}+3)}{(\sqrt{x^2+3}+2)}$$
$$=\lim_{x\to 1}\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+3}+2}$$
$$=\frac{3+3}{2+2}=\color{red}{\frac{3}{2}}$$
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