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I want to find the asymptote oblique of the following function for $x\rightarrow\pm\infty$

$$f(x)=\sqrt{x^2+3x}=\sqrt{x^2\left(1+\frac{3x}{x^2}\right)}\sim\sqrt{x^2}=|x|$$

For $x\rightarrow+\infty$ we have:

$$\frac{f(x)}{x}\sim\frac{|x|}{x}=\frac{x}{x}=1$$

which means that the function grows linearly.

$$f(x)-mx=\sqrt{x^2+3x}-x=x\left(\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}-1\right)\sim x\left(\frac {1}{2}\cdot\frac{3}{x}\right)=\frac{3}{2}$$

The oblique asymptote is $y=x+\frac 3 2$ which is correct. For $x\rightarrow-\infty$ we have:

$$\frac{f(x)}{x}=\frac{|x|}{x}=\frac{-x}{x}=-1$$

This means that

$$f(x)-mx=\sqrt{x^2+3x}+x=x\left(\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}+1\right)=x\left(\sqrt{1+\frac{3}{x}}+1\right)\sim x\cdot2\rightarrow -\infty$$

Which is not what my textbook reports ($-\frac{3}{2}$). Any hints on what I did wrong to find the $q$ for $x\rightarrow-\infty$?

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  • $\begingroup$ If $x < 0$ (or, better, $x\leq -3$), one has $\sqrt{x^2+3x} = |x| \sqrt{1+3/x} = -x \sqrt{1+3/x}$. $\endgroup$
    – Rigel
    Aug 18, 2018 at 9:17
  • $\begingroup$ For the second write $\sqrt{x^2+3x}+x=\dfrac{3x}{\sqrt{x^2+3x}-x}\to-\dfrac32$. $\endgroup$
    – Nosrati
    Aug 18, 2018 at 9:31

3 Answers 3

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The problem is in $$ f(x)-mx=\sqrt{x^2+3x}+x\underset{\begin{array}{c} \uparrow \\ \text{problem}\end{array}}{=}x\left(\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}+1\right)=x\left(\sqrt{1+\frac{3}{x}}+1\right)\sim x\cdot2\rightarrow -\infty $$ which should be $$ f(x)-mx=\sqrt{x^2+3x}+x=x\left(-\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}+1\right) $$ because we are near $-\infty$, so $\sqrt{x^2}=-x$.


In order to avoid such common mistakes, I suggest to change $x=-t$, when the limit for $x\to-\infty$ is involved: $$ \lim_{x\to-\infty}\bigl(\sqrt{x^2+3x}+x\bigr)= \lim_{t\to\infty}\bigl(\sqrt{t^2-3t}-t\bigr)= \lim_{t\to\infty}\frac{-3t}{\sqrt{t^2-3t}+t} $$ From here it should be clear.

Another strategy, in this case, is to set $x=-1/t$, that transforms the limit into $$ \lim_{t\to0^+}\left(\sqrt{\frac{1}{t^2}-\frac{3}{t}}-\frac{1}{t}\right)= \lim_{t\to0^+}\frac{\sqrt{1-3t}-1}{t} $$ which is a simple derivative: if $f(t)=\sqrt{1-3t}$, then $f'(t)=\frac{-3}{2\sqrt{1-3t}}$ and $$ \lim_{t\to0^+}\frac{\sqrt{1-3t}-1}{t}=f'(0)=-\frac{3}{2} $$

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This is not a different way, but is better I think $$f(x)=\sqrt{x^2+3x}=\sqrt{x^2+3x+(\frac32)^2-(\frac32)^2}=\sqrt{\left(x+\frac{3}{2}\right)^2-(\frac32)^2}\sim\left|x+\frac{3}{2}\right|$$ which gives both oblique asymptotes.

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$$f(x)=\sqrt{x^2+3x}=|x|\left(1+{3\over x}\right)^{1\over2}$$ which, as $x\rightarrow -\infty$, is equal to $$-x\left(1-{3\over{2x}}\right)^{1\over2} \sim -x\left(1+{3\over{2x}}\right)= -x -{3\over2}$$ using Taylor or Bernoulli

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  • $\begingroup$ Thanks. I haven't studied Taylor nor Bernoulli yet. $\endgroup$
    – Cesare
    Aug 18, 2018 at 9:26

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