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I have two functions, $f(x)=x$ and $g(x)=x-1$ $$\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=1$$ this means that when $x$ goes to infinity, those two function get closer and closer, So I think I can write it like this: $f(x)\approx g(x)$ as $x \rightarrow \infty$ but how come $$\int^{f(x)}_{g(x)} 1 \, dy \approx 0$$ we know for shout that this function must be 1. I just don't understand limit says that two functions are almost same, but the integral gives different answer.

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    $\begingroup$ Even more confusing: $f(x)=x+\ln x$ and $g(x)=x$. Then $f(x)\approx g(x)$ at $\infty$ but the integral $\int_{g(x)}^{f(x)}1\,dx=\ln x\to +\infty$. $\endgroup$ – A.Γ. Jul 18 '15 at 12:23
  • $\begingroup$ Just a comment on your notation. Asymptotic equivalence is usually denoted by ~, not $\approx$. $\endgroup$ – Muster Mark Jul 18 '15 at 12:46
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    $\begingroup$ The ratio having limit $1$ does not meet gets closer and closer. Graph $y=x$ and $y=x-1$. We get two parallel lines. $\endgroup$ – André Nicolas Jul 18 '15 at 13:04
  • $\begingroup$ Why do you write $\int^{f(x)}_{g(x)} 1 \, dy \approx 0$? I would not say that $1 \approx 0$. Are you sure you do not want to write $>$ or $\neq$ or even $\not\approx$ instead of $\approx$? $\endgroup$ – David K Jul 18 '15 at 14:33
  • $\begingroup$ This is exactly the problem with $\approx$ symbol. To me the statement $$\lim_{x \to \infty}\frac{f(x)}{g(x)} = 1$$ looks far more easier to grasp than the confusing $f(x) \approx g(x)$. One should never avoid rigor in favor of intuitive-ness/simplicity. Also this means that $f(x)/g(x)$ is close to $1$ and it does not mean that $f(x) - g(x)$ is close to $0$. $\endgroup$ – Paramanand Singh Jul 20 '15 at 10:19
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The fact that $$\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=1$$ does NOT mean that "when $x$ goes to infinity, those two function get closer and closer". It only means that the two functions get PROPORTIONALLY close, in the sense that $\vert f(x) - g(x)\vert$ gets PROPORTIONALLY small when compared to $g(x)$.

However, the absolute difference may $\vert f(x) - g(x)\vert$ may remain constant (or even increase).

In the case of $f(x)=x$ and $g(x)=x-1$ the absolute difference $\vert f(x) - g(x) \vert$ remains constant. So, in absolute terms, $f$ does not get closer to $g$.

In the case of $f(x)=x+\ln x $ and $g(x)=x$ the absolute difference $\vert f(x) - g(x)\vert$ increases but more slowly than $g(x)$. So, in absolute terms, $f$ even gets away from from $g$

Remark: Please note that it is easy to prove that $$\int^{f(x)}_{g(x)} 1 \, dy = f(x)-g(x)$$ so your questions regarding the integrals are actually questions about $f(x)-g(x)$.

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Whilst I wouldn't really use the notation $f(x) \approx g(x)$, since a functions tail having similar values at the tail doesn't mean they are anywhere close to the same function. Humouring your result, however, it seems that both results you've put forward are consistent with one another.

You get, from the limit of the ratio that $$f(x) \approx g(x)$$ as $x \to \infty$ and your integral agrees with this, since $$\int^{f(x)}_{g(x)} 1 \, \mathrm{d}y \approx 0 \implies \bigg[y\bigg]_{g(x)}^{f(x)}\implies f(x) - g(x) \approx 0 \implies f(x) \approx g(x).$$

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The integral says the same. Note that you can see that integral as a rectangle with sides 1 and $f(x)-g(x)$. The area of the rectangle is approximately 0, this means that $f(x)-g(x) \approx 0$, and thus $f(x)\approx g(x)$

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