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

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SUMMARY

The function g(x) = [3 + sin(1/x - 2)]/[1 + x^2] is proven to be bounded for all real numbers x, with specific bounds m and M established through analysis. The function f(x) is defined piecewise, being continuous for all x ≠ 2 and specifically at x = 2, utilizing the Squeeze Theorem to demonstrate continuity at that point. The discussion emphasizes the importance of understanding the behavior of g(x) as x approaches 2 and the implications for f(x).

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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)
 
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