I'm a Calculus 2 Student. But don't worry, this isn't a "do my HW" problem. I have a question about improper integrals.

I just realized something that's got me quite curious about the behaviors of infinite limits and I was hoping someone could explain what I'm observing.

If we're given the integral:

$$ \int_a^\infty [f(x)] dx $$

We're to take the limit of some number $M$ and solve the integral that way.

$$ \lim_{M\rightarrow\infty}\int_a^M [f(x)] dx $$    However, recalling what the definition of an integral is...

$$ \lim_{M\rightarrow\infty}[\lim_{N\rightarrow\infty} \sum_{k=1}^N[f(c_k) \Delta x] ] $$

What intrigues me is that we have two limits approaching infinity at the same time. One limit slicing the summation infinitely small, another expanding the scope of the summation infinitely long.

If something is becoming infinitely small and infinitely large at the same time, why doesn't it remain the same size? Does this imply $M$ is approaching infinity faster than $N$? Or does this mean the slices of $N$ (rather $\Delta x$ ) are "growing" with $M$ as it approaches infinity?


  • $\begingroup$ Nothing is "becoming infinitely small" nor "infinitely large". $\endgroup$ – Lord Shark the Unknown Feb 6 '18 at 4:25

These two limits are not happening at the same time. You're first doing the limit that slices things infinite small,producing a number, and then you're seeing what happens to this number as you make the interval larger and larger. These things happen in a very particular order.

For an intuitive description, think of it this way. Say the integral starts at $x=1$. Then the outer limit of your equations says that we fix $M$ at some finite value. At this fixed value, we chop up the function $f(x)$ into blocks, and do this until we get these "infinitely small blocks". We then take note of the final area and then raise the value of $M$ and restart this blocking procedure. The final answer is then the limit of this process as $M\rightarrow\infty$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.