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What I did was to see that the function is behaving as a straight line with gradient $3$ at infinity which implies that the function has an oblique asymptote as $x \to \infty$ but my testing portal says the answer is wrong and that instead, the function is infinite of order $1$ with respect to $x$. I think both of these are correct otherwise I must be confusing terminologies hereenter image description here

Botond
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M.Hamza Ali
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  • what does it mean "to be an infinite of order 1 at infinity" ? Is it $\mathcal O(x)$ ? – idm Feb 10 '18 at 10:22
  • i think it means that the order of the principal part at infinity is 1 – M.Hamza Ali Feb 10 '18 at 10:24
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    Consider $f(x)=3 x+ \log(x)$ which has $f'(x) = 3 +\frac1x \to 3$. Try to find the oblique asymptote – Henry Feb 10 '18 at 10:26
  • @M.HamzaAli Please, if you are ok, you can accept the answer and set it as solved. Thanks! cdn.sstatic.net/img/faq/faq-accept-answer.png – user Feb 12 '18 at 14:03

3 Answers3

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Note that in order to have a oblique asymptotes

  • $y=mx+n$ with $m,n\in \mathbb{R}$.

with

$$m=\lim_{x\rightarrow+\infty}\frac{f(x)}{x}$$

and

$$n=\lim_{x\rightarrow+\infty} (f(x)-mx)$$

both limit must exist.

Note that since the function is evantually strictly increasing $f(x)\to +\infty$ thus

$$\lim_{x\rightarrow+\infty}\frac{f(x)}{x}= \lim_{x\rightarrow+\infty}f'(x)=3$$

thus $f(x)=\mathcal O(x)$.

user
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The existence of an oblique asymptote is too strong. The derivative may approach 3 while the graph is not necessarily asymptotic to the line y=3x+b. It may oscillate and cross that line infinitely many times.

Mohammad Riazi-Kermani
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An example of a function on $\mathbb R$ with $\lim_{x\to\infty} f'(x) = 0$ but does not have a horizontal asymptote is $\log(1+x^2)$. If you prefer $f'(x) = 3$, try $$f(x) = 3x+\log(1+x^2).$$

Calvin Khor
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