Simple analysis continuity problem

economist1985
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Homework Statement


If f is a real function which is continuous at a element R and if f(a)<M for some M element of R, prove that there is an open interval I containing a such that f(x)<M for all x element of I.


Homework Equations


Extreme value theorem, intermediate value theorem, definition of continuity


The Attempt at a Solution


I have no idea how to solve this.
 
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Draw out the scenario. You have a continuous function with some point a on the x-axis, f(a) on the y-axis. Draw a horizontal line through the point f(a). We can think of M as a horizontal line somewhere above f(a). Let D be the distance between f(a) and M. If we can make f(x) within D of f(a), then M will be a bound for f(x) as well. But this is guaranteed by the continuity of f (epsilon-delta definition).
 
Thanks for the explanation. The problem is clearer now to me. What I still do not get is how the epsilon delta definition guarantees it. ?? Sorry if I can't follow through... I'm new to this.
 
Oh I think I got it. Thanks
 
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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