Is the Function f(x) Continuous at x=0 Given Its Definition and Behavior?

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SUMMARY

The function f(x) defined as f(x) = sin(x)sin(1/x) for x ≠ 0 and f(0) = 0 is continuous at x = 0. The limit as x approaches 0, lim_{x→0} sin(x)sin(1/x), is evaluated using the Squeeze Theorem. By establishing the bounds -|x| ≤ sin(x)sin(1/x) ≤ |x|, and noting that both limits of -|x| and |x| approach 0 as x approaches 0, it is confirmed that the limit equals 0. Therefore, f(x) is continuous at x = 0.

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kidia
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Using the inequalities -1 = < sin 1/x = < 1 for x not equal to 0 determine wether or not the function
f(x)= {sinxsin1/x x is not equal to 0
f(x)= {0 x=0

continuous at x=0
 
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Hmm...

Well, we obviously must find whether or not
[tex]\lim_{x\rightarrow0}\sin{x}\sin{\frac{1}{x}}=0[/tex]

Which is a bit of a challenge since [tex]\lim_{x\rightarrow\infty}\sin{\frac{1}{x}}[/tex] is undefined.
Looks like squeeze theorem time.
For all x, [tex]-|x|\leq\sin{x}\leq|x|[/tex] and [tex]|\sin{\frac{1}{x}}|\leq1[/tex].
Therefore, [tex]-|x|\leq\sin{x}\sin{\frac{1}{x}}\leq|x|[/tex] for all x.
Since [tex]\lim_{x\rightarrow0}-|x|=\lim_{x\rightarrow0}|x|=0, \lim_{x\rightarrow0}\sin{x}\sin{\frac{1}{x}}=0[/tex]

And thus it is continuous
 

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