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Does the function $$\frac{x}{\sqrt{1+x}}$$ have an oblique asymptote? If so, how do we find it? I thought about long dividing, but that wouldn't work, because You can't divide square roots.

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  • $\begingroup$ why not? $\frac{\sqrt{4}}{\sqrt{5}}$ exists.. $\endgroup$ – MonK Sep 30 '14 at 9:20
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You are not going to have an "oblique" asymptote like you are used to--it's not going to be a straight line. Instead the "oblique asymptote" is going to be a sideways parabola. This is because you have (essentially) $\frac{x}{\sqrt{x}} = \sqrt{x}$--so the asymtptote is going to be a square root--or quadratic:

$$ f(x) = \frac{x}{\sqrt{x + 1}} = x\frac{\sqrt{x + 1}}{x + 1} $$

Put another way what you have is: $\sqrt{x + 1} = y \rightarrow y^2 = x+1 \rightarrow x = y^2 - 1$:

$$ \left(y^2 - 1\left)\frac{y}{y^2}\right.\right. = \frac{y^2 - 1}{y} = y - \frac{1}{y} = \sqrt{x + 1} - \frac{1}{\sqrt{x + 1}} $$

This means that the "oblique" asymptote is $y = \sqrt{x + 1}$.

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    $\begingroup$ $y = \sqrt{x + 1}$ is a curvilinear asymptote. It is not oblique since an oblique asymptote is a line. $\endgroup$ – N. F. Taussig Sep 30 '14 at 9:32
  • $\begingroup$ @N.F.Taussig Hence why I put quotes around the term oblique...this function asymptotes to the function $f(x) = \sqrt{x}$ as I stated--it doesn't exactly asymptote to that, but it essentially does. $\endgroup$ – Jared Sep 30 '14 at 9:32
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    $\begingroup$ I knew you understood the question. I just wanted to make sure that the person who posted the question understood why you put the quotes there. $\endgroup$ – N. F. Taussig Sep 30 '14 at 9:33
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A necessary condition for the function $f$ to have an oblique asymptote at $\infty$ is that $$ \lim_{x\to\infty}\frac{f(x)}{x} $$ exists, is finite and is non zero. If the oblique asymptote exists, then this limit is its slope. It mustn't be zero, because otherwise it wouldn't be oblique in the first place.


If an oblique asymptote exists, with equation $y=mx+q$ ($m\ne0$), then, by definition, $$ \lim_{x\to\infty}(f(x)-mx-q)=0 $$ so also $$ \lim_{x\to\infty}\frac{f(x)-mx-q}{x}=0 $$ and then $$ \lim_{x\to\infty}\left(\frac{f(x)}{x}-m\right)=0 $$ because, of course, $\lim_{x\to\infty}\frac{q}{x}=0$.

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  • $\begingroup$ I don't see how an "oblique" asymptote differs from a $y = 0$ asymptote which you seem to single out. $\endgroup$ – Jared Sep 30 '14 at 9:52
  • $\begingroup$ @Jared A necessary condition for a horizontal asymptote is that $\lim_{x\to\infty}f(x)$ exists finite, which should be checked before trying for oblique asymptotes. $\endgroup$ – egreg Sep 30 '14 at 10:32
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    $\begingroup$ @Jared The usual example is $f(x)=\log x$, which hasn't a horizontal asymptote, but $\lim_{x\to\infty}f(x)/x=0$. $\endgroup$ – egreg Sep 30 '14 at 10:56
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HINT: $$\frac{x}{\sqrt{1+x}}=\frac{x+1-1}{\sqrt{x+1}}=\sqrt{1+x}-\frac{1}{\sqrt{1+x}}$$

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  • $\begingroup$ This is a good answer (don't get me wrong)...but I hope you (the OP) will take the time to understand my answer as well. $\endgroup$ – Jared Sep 30 '14 at 9:31
  • $\begingroup$ @Jared My answer is purely a hint,yours is a complete answer :) $\endgroup$ – kingW3 Sep 30 '14 at 9:37
  • $\begingroup$ I agree, hence why I upvoted your answer as it agreed with mine...I would suggest the OP and others to do the same since it verifies the result (regardless of who came first). $\endgroup$ – Jared Sep 30 '14 at 9:38

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