Evaluating $\lim_{x\to1}{\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}}$ without L'Hospital's Theorem

Question

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?

Answer 1

Notice, $$\lim_{x\to 1}\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}$$ $$=\lim_{x\to 1}\frac{(\sqrt{x^2+3}-2)(\sqrt{x^2+3}+2)}{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}\cdot \frac{(\sqrt{x^2+8}+3)}{(\sqrt{x^2+3}+2)}$$ $$=\lim_{x\to 1}\frac{x^2+3-4}{x^2+8-9}\cdot \frac{(\sqrt{x^2+8}+3)}{(\sqrt{x^2+3}+2)}$$ $$=\lim_{x\to 1}\frac{(x^2-1)}{(x^2-1)}\cdot \frac{(\sqrt{x^2+8}+3)}{(\sqrt{x^2+3}+2)}$$ $$=\lim_{x\to 1}\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+3}+2}$$

$$=\frac{3+3}{2+2}=\color{red}{\frac{3}{2}}$$

Answer 2

Rationalize the denominator.I.e multiply the numerator and the denominator by $\sqrt{x^2+8}+3$...

Answer 3

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 $$\lim_{x\to1}\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+3}+2} = \frac{3}{2}$$

Answer 4

\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 $(a-b)(a+b)=a^2-b^2$.

Answer 5

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.

Answer 6

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

Answer 7

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?

$\lim_{x\to1}{\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3} * \frac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3}} = lim_{x\to1}{\frac{(\sqrt{x^2+3}-2 )* (\sqrt{x^2+8}+3)}{x^2-1}} = lim_{x\to1}{\frac{(\sqrt{x^2+3}-2 )* (\sqrt{x^2+8}+3)}{x^2-1}*\frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2}} = lim_{x\to1}{\frac{(x^2-1 )* (\sqrt{x^2+8}+3)}{x^2-1}*\frac{1}{\sqrt{x^2+3}+2}} = = lim_{x\to1}{\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+3}+2}} = \frac{\sqrt{1^2+8}+3}{\sqrt{1^2+3}+2} = \frac{6}{4} = \frac{3}{2}$