0
$\begingroup$

Find all the horizontal asymptotes of $f(x)=\frac{x^3+2}{x^3+\sqrt{4x^6+3}}$

I am completely lost on this one, when I work out the limits I get Horizontal Asymptotes of $-\frac{1}{3}$, which I have already been told is wrong

  • 2
    only wrong by a minus sign ! – 2017-02-18
  • 1
    Factorize $x^3$ both at the numerator and denominator... – 2017-02-18
  • 2
    Use $\sqrt{4x^5 + 3} \sim \lvert 2x^3\rvert$. Distinguish between the cases $x\to +\infty$ and $x\to -\infty$. – 2017-02-18
  • 0
    To my understanding there should be 2 asymptotes – 2017-02-18

2 Answers 2

1

Note that $$\frac{\sqrt{4x^6+3}}{x^3}=\begin{cases}\sqrt{4+\frac3{x^6}}\text{ if }x>0\\-\sqrt{4+\frac3{x^6}}\text{ if }x<0\end{cases}$$

Then $$\lim_{x\to\infty}\frac{x^3+2}{x^3+\sqrt{4x^6+3}}=\lim_{x\to\infty}\frac{1+\frac2{x^3}}{1+\sqrt{4+\frac3{x^6}}}=\frac13$$

$$\lim_{x\to-\infty}\frac{x^3+2}{x^3+\sqrt{4x^6+3}}=\lim_{x\to-\infty}\frac{1+\frac2{x^3}}{1-\sqrt{4+\frac3{x^6}}}=-1$$

  • 0
    I am kind of lost on how to bring everything under the radical – 2017-02-18
  • 0
    If $a,b\ge 0$ then $a\sqrt b=\sqrt{a^2b}$. – 2017-02-18
  • 0
    Ok i think i see that a little bit more... to bring $x^3$ in the radical we square it and distribute. And the reason the negative stays is because it is to an odd power so the negative maintains as $x \rightarrow -\infty$? Is that right? – 2017-02-18
  • 0
    If $a<0$ and $b>0$ then $a\sqrt b=-|a|\sqrt b=-\sqrt{|a|^2b}=-\sqrt{a^2b}$. – 2017-02-18
1

Hint : Divide numerator and denominator by $x^3$ and use $$\frac{\sqrt{4x^6+3}}{2x^3}=\sqrt{1+\frac{3}{4x^6}}$$