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?
 A: Notice, $$\lim_{x\to 1}\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}$$
$$=\lim_{x\to 1}\frac{(\sqrt{x^2+3}-2)(\sqrt{x^2+3}+2)}{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}\cdot \frac{(\sqrt{x^2+8}+3)}{(\sqrt{x^2+3}+2)}$$
$$=\lim_{x\to 1}\frac{x^2+3-4}{x^2+8-9}\cdot \frac{(\sqrt{x^2+8}+3)}{(\sqrt{x^2+3}+2)}$$
$$=\lim_{x\to 1}\frac{(x^2-1)}{(x^2-1)}\cdot \frac{(\sqrt{x^2+8}+3)}{(\sqrt{x^2+3}+2)}$$
$$=\lim_{x\to 1}\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+3}+2}$$
$$=\frac{3+3}{2+2}=\color{red}{\frac{3}{2}}$$
A: Rationalize the denominator.I.e multiply the numerator and the denominator by $\sqrt{x^2+8}+3$...
A: Rationalise both numr and denr
$$\frac{x^2+3-4}{x^2 + 8 -9}\times\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+3}+2}$$
apply limit
$$\lim_{x\to1}\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+3}+2} = \frac{3}{2}$$
A: \begin{align*}
\lim_{x\to1}\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}
&=\lim_{x\to1}\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\to1}\frac{x^2-1}{x^2-1}\cdot\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+3}+2}\\
&=\frac32
\end{align*}
using $(a-b)(a+b)=a^2-b^2$.
A: Hint: 
$$\begin{align}
\lim_{x\to1}{\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}}&=\lim_{x\to1}\frac{\sqrt{x^2+3}-2}{x-1}\frac{x-1}{\sqrt{x^2+8}-3}\\&=\lim_{x\to1}\frac{\sqrt{x^2+3}-2}{x-1}\left(\frac{\sqrt{x^2+8}-3}{x-1}\right)^{-1}.
\end{align}$$
Notice that it would be sufficient to calculate two derivatives at $x=1$ to evaluate the limit.
A: 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?
$\lim_{x\to1}{\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3} * \frac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3}} = lim_{x\to1}{\frac{(\sqrt{x^2+3}-2 )* (\sqrt{x^2+8}+3)}{x^2-1}} = lim_{x\to1}{\frac{(\sqrt{x^2+3}-2 )* (\sqrt{x^2+8}+3)}{x^2-1}*\frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2}} = lim_{x\to1}{\frac{(x^2-1 )* (\sqrt{x^2+8}+3)}{x^2-1}*\frac{1}{\sqrt{x^2+3}+2}} = = lim_{x\to1}{\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+3}+2}} = \frac{\sqrt{1^2+8}+3}{\sqrt{1^2+3}+2} = \frac{6}{4} = \frac{3}{2}$
A: $$\lim_\limits{x\to1} \frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}$$
$$=\lim_\limits{x\to1} \left[\frac{\frac{\sqrt{x^2+3}-2}{x^2-1}}{\frac{\sqrt{x^2+8}-3}{x^2-1}}\right]$$
$$=\lim_\limits{x\to1} \left[\frac{\frac{\sqrt{x^2+3}-2}{x^2+3-2}}{\frac{\sqrt{x^2+8}-3}{x^2+8-9}}\right]$$
$$=\frac{\lim_\limits{x^2+3\to4} \frac{\sqrt{x^2+3}-\sqrt4}{(x^2+3)-4}}{\lim_\limits{x^2+8\to9} \frac{\sqrt{x^2+8}-\sqrt9}{(x^2+8)-9}}$$
$$=\frac{\frac{1}{2}4^{\frac{1}{2}-1}}{\frac{1}{2}9^{\frac{1}{2}-1}}$$
$$=\frac{\frac{1}{4}}{\frac{1}{6}}$$
$$=\frac{3}{2}$$
