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I'm reading Stewart's calculus book and I faced with an intuitive issue.

Intuitively, I don't understand why since both graphs looks like the same up to a zoom in/out. Could someone explain to me this intuitively? For me both should be convergent in this context.

It's difficult to grasp this intuitively because it's like both functions are approaching to zero without the area of the second one approaching to zero, how is that possible?

user42912
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    By that standard, presumably any decreasing function with an asymptote at zero would look "the same up to a zoom in/out" – Ben Grossmann Jan 06 '20 at 10:56
  • @Omnomnomnom good point, maybe so. But this not answer my question. – user42912 Jan 06 '20 at 11:02
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    Isn't the answer within the text you posted? "Note that... $\frac{1}{x^2}$ approches $0$ faster... The values of $\frac{1}{x}$ don't decrease fast enough..." – Déjà vu Jan 06 '20 at 11:03
  • @RingØ no. Intuitively, faster doesn't mean anything in this context since we are talking about infinity. If the graph stopped somewhere, I could understand the difference between both, but both goest to infinity. – user42912 Jan 06 '20 at 11:07
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    @RingØ Note both areas are decreasing because both are approaching to zero – user42912 Jan 06 '20 at 11:08
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    @user42912: It's the other way around! If the graphs stopped, the areas would simply both be finite, regardless of how "fast" the functions decrease. It's precisely when you have an interval extending out to infinity that it's interesting to ask whether the functions decrease "fast" enough to make the total area finite or not. – Hans Lundmark Jan 06 '20 at 11:34

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They don’t look the same up to a zoom in zoom out! The curvature is different - you can see that the factor of the vertical stretch from left graph to right graph is not constant but grows proportional to $x$: the right most values look like they are maybe 400% bigger, but near $x=1$ you can see the stretch is <100%. So it doesnt look like just zooming.

Ben
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Intuitively, faster doesn't mean anything in this context since we are talking about infinity. If the graph stopped somewhere,

You don't ever need to directly talk about infinity. The definition of the improper integral $\int_0^\infty$ is by a limiting process; you essentially get the answer by first stopping somewhere, and then asking what happens if you stop at a larger number, over and over. And you can check $$ \int_1^C \frac{dx}{x^2} = 1-\frac1C$$ which is controllable; its never larger than 1, no matter the value of $C$, whereas $$ \int_1^C \frac{dx}x = \log C$$ can be made as large as you want. (This is essentially a full proof once you know the right definitions and note that $\frac{1}x>0$ so that the integral is increasing in $C$.)

Calvin Khor
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  • I understand that, my only problem is I can't understand that visually in the graph. It's confusing a function approaching to zero whose area isn't approaching to anywhere. – user42912 Jan 06 '20 at 12:06
  • Why are you using your eyes to do the integration, and then taking a limit? If you graphed $\log C$ instead you'd have a hard time arguing. And further, if you have any sensible definition of an integral $\int_1^\infty$, you better have $$ f>0 \implies \int_1^C f(x) dx \le \int_1^\infty f(x) dx$$ so the finite graph already produces great lower bounds @user42912 – Calvin Khor Jan 06 '20 at 12:15