find $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3} \ -\sqrt{2x+2}}$ without l'hospital rule

Question

EDITED VERSION

find $$\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3} \ -\sqrt{2x+2}}$$ without l'hospital rule.

using l'hospital rule, you'll have: $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3} \ -\sqrt{2x+2}} = -6$.

I can show my attempt but it's pointless as the result i achieved $+\infty$, which is very wrong. is there an strategy trying to calculate a limit without l'hospital rule.

Answer 1

The expression tends to $$\frac{2\cdot1-\sqrt{1^2+3}}{\sqrt{1+3}+\sqrt{2\cdot1+2}}=0.$$

You may not use L'Hospital here. (Or is there a typo in the question ?)

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Update:

$$\frac{2x-\sqrt{x^2+3}}{\sqrt{x+3}-\sqrt{2x+2}}=\frac{\sqrt{x+3}+\sqrt{2x+2}}{2x+\sqrt{x^2+3}}\cdot \frac{4x^2-x^2-3}{x+3-2x-2}\to\frac{4}{4}(-6).$$

Answer 2

I will assume that you meant$$\lim_{x\to1}\frac{2x-\sqrt{x^2+3}}{\sqrt{x+3}-\sqrt{2x+2}}.$$Multiplying the numerator and the denominator by $\sqrt{x+3}+\sqrt{2x+2}$, this becomes$$\lim_{x\to1}\frac{\left(2x-\sqrt{x^2+3}\right)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{1-x}.$$So, define $f(x)$ as $2x-\sqrt{x^2+3}$ and then the limit that you're after is just$$-f'(1)\left(\sqrt{1+3}+\sqrt{2+2}\right)=-4f'(1)$$(which is $-6$ indeed).

Answer 3

Multiplying the top and the Bottom by $ \left(2x+\sqrt{x^2+3}\right)\left(\sqrt{x+3}+\sqrt{2x+2}\right) $, gives the following :

\begin{aligned} \lim_{x\to 1}{\frac{2x-\sqrt{x^2+3}}{\sqrt{x+3}-\sqrt{2x+2}}}&=\lim_{x\to 1}{\frac{-3\left(x+1\right)\left(\sqrt{2x+2}+\sqrt{x+3}\right)}{\sqrt{x^{2}+3}+2x}}\\ &=\frac{-3\left(1+1\right)\times\left(\sqrt{2\times 1+2}+\sqrt{1+3}\right)}{\sqrt{1^{2}+3}+2\times 1}\\ \lim_{x\to 1}{\frac{2x-\sqrt{x^2+3}}{\sqrt{x+3}-\sqrt{2x+2}}}&=-6 \end{aligned}

Answer 4

Starting from J.C.Santos result

$$\lim_{x\to1}\frac{\left(2x-\sqrt{x^2+3}\right)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{1-x}.$$

I'd multiply num and den by $\left(2x+\sqrt{x^2+3}\right)$ to get

$$\lim_{x\to1}\frac{\left(4x^2-x^2-3\right)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{(1-x)\left(2x+\sqrt{x^2+3}\right)}.$$ that is $$\lim_{x\to1}\frac{3(x+1)(x-1)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{(1-x)\left(2x+\sqrt{x^2+3}\right)}.$$ and then $$-\lim_{x\to1}\frac{3(x+1)\left(\sqrt{x+3}+\sqrt{2x+2}\right)}{\left(2x+\sqrt{x^2+3}\right)}=-\frac{24}{4}=-6$$