How do I calculate $$\lim_{x\to-\infty}2x+\sqrt{4x^2+x}?$$ I've gotten "close" many times but I just can't figure it out, I always get stuck with zero division or something similar.
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Also, how do I format the formulas nicely? – Mahi Sep 26 '17 at 07:54
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See this thread for a quick guide on how to format mathematics on math.SE and its sister sites. – Arthur Sep 26 '17 at 07:55
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@Arthur Thanks mate, bookmarked that for future cases! – Mahi Sep 26 '17 at 07:56
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1The idea is that when $x$ is large, $x^2$ is very big compared to $x$, so for $x$ large and negative, $x$ is negligible compared to $4x^2$, so $\sqrt{4x^2+x}\approx\sqrt{4x^2}=2|x|$. If you want to make this argument rigorous, there are ways... – Teddy38 Sep 26 '17 at 07:58
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See https://math.stackexchange.com/questions/2214526/find-lim-x-to-infty-sqrtx29x3/2214533#2214533 – lab bhattacharjee Sep 27 '17 at 05:14
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Define $t:=-x$. Then, \begin{align*} &\lim_{x \to -\infty}2x+\sqrt{4x^2+x}=\lim_{t \to \infty} \left( -2t+\sqrt{4t^2-t} \right)\\ &=\lim_{t \to \infty} \left( -2t+\sqrt{4t^2-t} \right)\cdot\frac{2t+\sqrt{4t^2-t}}{2t+\sqrt{4t^2-t}} =\lim_{t \to \infty}\frac{-t}{2t+\sqrt{4t^2-t}} \\ &=\lim_{t \to \infty}\frac{-1}{2+\frac{\sqrt{4t^2-t}}{t}}=\lim_{t \to \infty}\frac{-1}{2+\sqrt{\frac{4t^2-t}{t^2}}}=\lim_{t \to \infty}\frac{-1}{2+\sqrt{\frac{4t^2-t}{t^2}}}\\ &=\lim_{t \to \infty}\frac{-1}{2+\sqrt{4-(1/t)}}=\frac{-1}{2+\sqrt{4}}=-1/4. \end{align*}
sharpe
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While all answers were probably right, this is the one that clicked for me and I understood everything :) Thanks mate! – Mahi Sep 26 '17 at 08:30
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\begin{align}\lim_{x\to-\infty}2x+\sqrt{4x^2+x}&=\lim_{x\to-\infty}\frac{\left(2x+\sqrt{4x^2+x}\right)\left(2x-\sqrt{4x^2+x}\right)}{2x-\sqrt{4x^2+x}}\\&=\lim_{x\to-\infty}\frac{-x}{2x-\sqrt{4x^2+x}}\\&=-\lim_{x\to-\infty}\frac1{2+\sqrt{4+\frac1x}}\text{ (because $x<0$)}\\&=-\frac14.\end{align}
José Carlos Santos
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$$\begin{align*} \lim_{x \rightarrow -\infty} 2x + \sqrt{4x^2 + x} & = \lim_{x \rightarrow \infty} \sqrt{4x^2 - x} - 2x = \\ & = \lim_{x \rightarrow \infty} \frac{(\sqrt{4x^2 - x} - 2x)(\sqrt{4x^2 - x} + 2x)}{\sqrt{4x^2 - x} + 2x} = \\ & = \lim_{x \rightarrow \infty} -\frac{x}{\sqrt{4x^2 - x} + 2x} = \boxed{-\frac{1}{4}} \end{align*}$$
cip999
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Same here, I don't get how the last step happens and you end up with $-\frac{1}{4}$ – Mahi Sep 26 '17 at 08:19
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Making the substitution $u = 1/x$, we aim to find $$\lim_{u\to 0^{\large{-}}} \left(2/u + \sqrt{4/u^2 + 1/u}\ \right).$$ When $u < 0$, we have $\sqrt{u^2} = -u$, so $\sqrt{4/u^2 + 1/u} = \sqrt{(4+u)/u^2}=-\sqrt{4 + u}/u$. Hence, we obtain $$\lim_{u\to 0^{\large{-}}} \frac{2-\sqrt{4 + u}}{u}.$$ The numerator is $0$ at $u=0$. So let $f(u) = 2 - \sqrt{4+u}$. If we use the definition of the derivative, we see that our limit can be written as $$\lim_{u\to 0^{\large{-}}}\frac{f(u) - f(0)}{u} = f'(u)\Big|_{u=0}$$ provided the derivative exists. Applying the standard rules of differentiation, $f'(u) = -1/(2\sqrt{4+u})$. This equals $-1/4$ at $u=0$. Hence, $$\lim_{x\to-\infty} \left(2x + \sqrt{4x^2 +x}\right) = -1/4.$$
Eric M. Schmidt
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Note that for $x\to -\infty$, $4x^2+x=(2x+\frac14)^2-\frac{1}{16} \sim (2x+\frac14)^2.$ Hence: $$\lim_{x\to-\infty} \left(2x+\sqrt{4x^2+x}\right)=\lim_{x\to-\infty}\left(2x-\left(2x+\frac14\right)\right)=-\frac14.$$
farruhota
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