Continuity in Analysis: Finding Continuous Functions

ryanj123
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Homework Statement



Find sets of all x on which the following functions are continuous
using any theorems available.

When the phrase "any thms. available" is used, we are only at a stage in my beginning analysis course where we've learned up to continuity, limits, convergence/divergence, circular functions, etc. Not much beyond that, so the proof I'm trying to construct needs to fall within these limits.

Homework Equations



i) 1/(sqrt(x^2+2x+2))

ii) (x(x-1))/(x^2+2x-2)

iii) sec(x^2)

The Attempt at a Solution



So I've graphed these things to better see the continuity.

For i) obviously when the sqrt is greater than or equal to 0 is it defined.

For ii) the function is discontinuous by means of the quadratic formula at x=-sqrt(3)-1, and x=sqrt(3)-1.

For iii) this function can be represented as 1/(cos(x^2)) which is discontinuous when cos(x^2)=0.

So I understand these areas of discontinuity and continuity but I'm not sure how to formulate it exactly into an argument involving:

epsilon>0 , there exists delta>0 such that:

lf(x)-f(a)l<epsilon => lx-al<delta
 
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Try showing that each of them are continuous whenever they're defined.
 
The problem says "using any theorems available" so I see no reason to go back to the "epsilon-delta" definitions.

You surely know things like "if f(x) and g(x) are both continuous at x= a then so are f(x)+ g(x) and f(x)g(x)", "If f(x) and g(x) are both continuous at x= a and g(a) is not 0 then f(x)/g9x) is continuous at x= a", "If g(x) is continuous ata x= a and f(x) is continuous at x= f(a) then f(g(x)) is continuous at a", "any polynomial is continuous for all a", "\sqrt{x}" is continuous for all non-negative a", and "cos(x) is continuous for all a". Putting those together will give the correct continuity for each of these functions.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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