Is there a general way of finding this. Usually what I find on the internet is dividing the function by ax + b but I can't seem to make it work
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$\begingroup$ For $x\to\infty$ we have$$xe^{1/x}=x\left(1+\frac1x+o\left(\frac1x\right)\right)=x+1+o(1)$$ $\endgroup$– Peter ForemanCommented Jul 3, 2020 at 19:24
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$\begingroup$ I don't undersand. Where did the o(1/x) come from? $\endgroup$– Eduardo BrancoCommented Jul 3, 2020 at 19:31
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$\begingroup$ An application of Taylor's theorem on the function $e^{1/x}$. I could have also written $O(1/x^2)$ instead. $\endgroup$– Peter ForemanCommented Jul 3, 2020 at 19:33
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$\begingroup$ Fair enough. Is there a simpler way though? $\endgroup$– Eduardo BrancoCommented Jul 3, 2020 at 19:40
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$\begingroup$ See the answer below. Although in order to calculate these limits one is going to have to use the above result in some way to study the functions behaviour. $\endgroup$– Peter ForemanCommented Jul 3, 2020 at 19:42
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1 Answer
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Hint: Use differential geometry!
The oblique asymptotes have the equation:
$$y=kx+b, \space \text{ with } \space \space k = \lim_{x \to \infty} \frac{f(x)}{x}, \space \space b = \lim_{x \to \infty} [f(x) - kx].$$
answered Jul 3, 2020 at 19:36
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$\begingroup$ This is right. Although it didn't help me a lot because the finding b is so hard for me in this case. But now I know how to do it $\endgroup$ Commented Jul 3, 2020 at 20:06