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May I know how can I calculate the following expression?

$$ \lim\limits_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2} $$

From calculator, the answer is $-\sqrt{6}$ , my approach is by using dividing numerator and denominator by using $x$. Which is,

$$ \lim\limits_{x\to-\infty}\frac{\frac{\sqrt{6x^2 - 2}}{x}}{\frac{x+2}{x}} = \lim\limits_{x\to-\infty}\frac{{\sqrt{\frac{6x^2 - 2}{x^2}}}}{\frac{x+2}{x}}=\sqrt{6} $$

My answer is $\sqrt{6}$, is my working wrong or there are actually another approach? Thank you.

Juniven Acapulco
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Ritsuji
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5 Answers5

4

Try looking at it this way: $$\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2}\cdot \sqrt{6-\frac{2}{x^2}}}{x+2}$$ Then since $x \rightarrow -\infty$ we have $x<0$ thus $\sqrt{x^2}=|x| = -x$ and we have $$\lim_{x\rightarrow -\infty} \frac{-x\cdot \sqrt{6-\frac{2}{x^2}}}{x+2}=\lim_{x\rightarrow -\infty} \frac{-\sqrt{6-\frac{2}{x^2}}}{1+\frac{2}{x}}=-\sqrt6$$

Sheheryar Zaidi
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If $x\lt 0$, then $$\frac{\sqrt{6x^2-2}}{x}\not=\sqrt{\frac{6x^2-2}{x^2}}$$ because the left-hand-side is negative and the right-hand-side is positive.

Since $-x\gt 0\Rightarrow -x=\sqrt{(-x)^2}=\sqrt{x^2}$, we have

$$\frac{\sqrt{6x^2-2}}{x}=\frac{\sqrt{6x^2-2}}{-(-x)}=\frac{\sqrt{6x^2-2}}{-\sqrt{(-x)^2}}=-\sqrt{\frac{6x^2-2}{(-x)^2}}=-\sqrt{\frac{6x^2-2}{x^2}}.$$

mathlove
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it is $\frac{|x| \sqrt{6-\frac{2}{x^2}}}{x(1+\frac{2}{x})}=-\frac{\sqrt{6-\frac{2}{x^2}}}{1+\frac{2}{x}}$ and the searched limit is $-\sqrt{6}$

Dr. Sonnhard Graubner
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$$\lim_{x\to -\infty} \frac{\sqrt{6x^2-2}}{x+2}$$ First note that as $x\to -\infty$, then $x+2\lt 0$ and $$ x+2=-|x+2|=-\sqrt{(x+2)^2} $$ Therefore $$ \lim_{x\to -\infty} \frac{\sqrt{6x^2-2}}{x+2} = \lim_{x\to -\infty} \frac{\sqrt{6x^2-2}}{-\sqrt{(x+2)^2}} $$ $$ = -\lim_{x\to -\infty} \sqrt{\frac{6x^2-2}{x^2+4x+4}} = -\lim_{x\to -\infty} \sqrt{\frac{6-\frac{2}{x^2}}{1+\frac{4}{x}+\frac{4}{x^2}}} $$ $$ =-\sqrt{\frac{6-0}{1+0+0}}=-\sqrt{6} $$

k170
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0

It is all too easy (for me) to make a sign error. Knowing that, I would probably try to scurry to safety by letting $x=-t$. We have $$\lim_{x\to -\infty}\frac{\sqrt{6x^2-2}}{x+2}=\lim_{t\to \infty}\frac{\sqrt{6t^2-2}}{-t+2}=\lim_{t\to \infty}-\frac{\sqrt{6t^2-2}}{t-2}.$$ Now do the usual sort of thing. Since we are thinking of $t$ as large positive, our chance of error decreases.

André Nicolas
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