When can you take the limit of both sides?

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Homework Statement



A couple questions that have been building up here. I usually don't see them explicitly answered in the text.

When can you take the limit of both sides of an equation? And can you get rid of the limit from both sides of an equation?

How about taking the derivative or integral of both sides?

Can you ever take the absolute value of both sides?

What about taking the sup or inf of both sides?


These questions should help me a little bit.
 
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You can always do any of the above on both sides. Whether it exists on each side or whether they are equal on both sides depends completely on the problem. Sorry if that's not much an answer, but it's not a very good question.
 
Sorry about the question. I'm worse with words than math, and you know about my math skills Dick.

I see that a law like lim[ f(x) + g(x)] = f+g, might confirm this. I usually take this law in one direction -->. But going the other direction atleast partly justifies what I was asking.
 
If f and g are continuous at x0 then, sure, lim x->x0 f(x)+g(x)=f(x0)+g(x0) from either side. If they aren't then all bets are off. It depend exactly on what f and g are and what exactly you are trying to do. There is no one-size-fits-all answer to your question.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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