Bound Function: Showing Continuity at All x ≠ 2 & x = 2

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(a) Show that the function g(x) =[3 + sin(1/x-2)]/[1 + x^2] is bounded.
This means to find real numbers m; M is an lR such that m ≤ g(x) ≤ M for
all x is an lR (and to show that these inequalities are satisfied!).

(b) Explain why the function:

f(x) = { [x-2] [3 + sin(1/x-2)]/[1 + x^2] , if x ≠ 2,
{ 0 , if x = 2.

is continuous at all x ≠ 2.

(c) Show that the function f(x) in Part (b) is continuous at x = 2. [Hint: Use
Part (a) and the Squeeze Theorem.]
 
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how about considering g(x) = u(x)/v(x) and each of the behaviours of those functions

note that g(x) will get big whenever u(x)>>v(x)
 
Welcome to PF;
How about showing us your attempt at the problem? ... that way we can target our assistance to where you need it most.
 
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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