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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?

N. F. Taussig
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ILoveChess
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  • Do you have any idea what kind of things you are supposed to use here? since you suggest that you need to use things presented in lecture, and I'm not sure what kinda things are covered in your lecture... – user160738 Feb 03 '16 at 15:00

7 Answers7

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

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

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

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

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

Cm7F7Bb
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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.

Workaholic
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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?

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

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

SchrodingersCat
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