<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] Here is a same solution:How can I evaluate $\lim_{x\to1}\frac{\sqrt{5-x}-2}{\sqrt{2-x}-1}$ without invoking l'Hôpital's rule?
$\lim_{x\to1}{\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3} * \frac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3}} = lim_{x\to1}{\frac{(\sqrt{x^2+3}-2 )* (\sqrt{x^2+8}+3)}{x^2-1}} = lim_{x\to1}{\frac{(\sqrt{x^2+3}-2 )* (\sqrt{x^2+8}+3)}{x^2-1}*\frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2}} = lim_{x\to1}{\frac{(x^2-1 )* (\sqrt{x^2+8}+3)}{x^2-1}*\frac{1}{\sqrt{x^2+3}+2}} = = lim_{x\to1}{\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+3}+2}} = \frac{\sqrt{1^2+8}+3}{\sqrt{1^2+3}+2} = \frac{6}{4} = \frac{3}{2}$
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