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I've been working on this problem for a while now, but I can't solve it

$$\lim\limits_{x\to-\infty}{\sqrt{4x^2+x+7}+2x}$$

I've tried multiplying by

$$\frac{\sqrt{4x^2+x+7}-2x}{\sqrt{4x^2+x+7}-2x}$$

but I didn't get it. Am I missing something really obvious? Can someone help me with this question?

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6 Answers 6

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$$ \begin{aligned} \lim_{x\to-\infty}\left(\sqrt{4x^2+x+7}+2x\right) &= \lim_{x\to-\infty}\frac{4x^2+x+7-4x^2}{\sqrt{4x^2+x+7}-2x}\\ & =\lim_{x\to-\infty}\frac{x+7}{\sqrt{4x^2+x+7}-2x}\\ &=\lim_{x\to-\infty}\frac{-1-\frac{7}{x}}{\sqrt{4+\frac{1}{x}+\frac{7}{x^2}}+2}\\ &=-\frac{1}{4} \end{aligned} $$

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  • $\begingroup$ How do you know that $\sqrt{4}$ isn't $-2$? $\endgroup$ Commented Feb 20, 2020 at 13:09
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    $\begingroup$ @NoLand'sMan, what do you mean? The square root is always non-negative. $\endgroup$
    – LHF
    Commented Feb 20, 2020 at 13:10
  • $\begingroup$ $(-2)^2$ is also $4$ $\endgroup$ Commented Feb 20, 2020 at 13:15
  • $\begingroup$ @NoLand'sMan, I think you confuse solving a equation of the form $x^2=4$ with the definition of the square root function. Notice that $f(x)=x^2$ and $f(x)=\sqrt{x}$ are not the inverse of each other since they have different definition domains. $\endgroup$
    – LHF
    Commented Feb 20, 2020 at 13:18
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Hint: Multiplying by $\ \frac{\sqrt{4x^2+x+7}-2x}{\sqrt{4x^2+x+7}-2x}\ \left(=1\right)\ $ shows that \begin{align} \sqrt{4x^2+x+7}+2x&=\frac{x+7}{\sqrt{4x^2+x+7}-2x}\\ &=\frac{-1+\frac{7}{|x|}}{\sqrt{4-\frac{1}{|x|}+ \frac{7}{x^2}}+2}\ \ \text{for }\ x<0\ . \end{align} Can you see what the limit of this last expression is?

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To find $\lim_{x \to -\infty} \frac {x+7} {\sqrt {4x^{2}+x+7} -2x}$ divide the numerator and the denominator by $-x$. The limit is $-1/4$.

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\begin{gather*} \lim _{x\rightarrow -\infty }\sqrt{4x^{2} +x+7} -2x\ =\ \lim _{x\rightarrow -\infty }\sqrt{4x^{2}( 1+\frac{x}{4x^{2}} +\frac{7}{4x^{2}}} -2x\ =\ \ \\ \lim _{x\rightarrow -\infty } 2x\sqrt{( 1+\frac{1}{4x} +\frac{7}{4x^{2}}} -2x\ \ Apply\ Binomial\ theorem\\ \lim _{x\rightarrow -\infty } 2x\left( 1+\frac{1}{8x} +\frac{7}{8x^{2}}\right) -2x\ =\ \frac{1}{4}\\ \end{gather*}

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  • $\begingroup$ I think you mean $-\frac14$ $\endgroup$
    – nyz
    Commented Feb 25, 2020 at 3:56
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Set $x=-\dfrac1h,\implies h\to0^+$

$$\sqrt{4x^2+x+7}=\sqrt{\dfrac{4-h+7h^2}{h^2}}=\dfrac{\sqrt{4-h+7h^2}}h$$

So, we need $$\lim_{h\to0^+}\dfrac{\sqrt{4-h+7h^2}-2}h=\lim_{h\to0^+}\dfrac{4-h+7h^2-2^2}{h(\sqrt{4-h+7h^2}+2)}=\dfrac{-1}{\sqrt4+2}$$

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$$L=\lim_{x \rightarrow \infty} \sqrt{4x^2+x+7}+2x$$ $$\implies L= \lim_{x \rightarrow -\infty} |2x| \sqrt{1+\frac{1}{4x}+\frac{7}{4x^2}}+2x$$ $$\implies L=-2x\left(1+\frac{1}{8x}+\frac{7}{8x^2}+..+O(1/x^2)\right)+2x$$ $$L=-\frac{1}{4}.$$

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