Find the Limit of g(x)sinx as x Approaches 0: Homework Help & Equations

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SUMMARY

The limit of g(x)sin(x) as x approaches 0 is definitively 0. The function g(x) is defined as 1 for rational x and 0 for irrational x. The analysis shows that |g(x)sin(x)| is bounded by |sin(x)|, which approaches 0 as x approaches 0. Therefore, the limit can be established using the Squeeze Theorem, confirming that lim g(x)sin(x) = 0 as x approaches 0.

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Homework Statement


g(x) = 1 if x is rational, 0 if x is irrational.

Determine the limit, or prove it doesn't exist.

lim g(x)sinx
x->0

Homework Equations


The Attempt at a Solution


I said the limit is 0.

0 < |g(x)sinx| < |sinx|

lim |g(x)sinx|= 0
x->0
-|sinx| < g(x)sinx < sinx

lim g(x)sinx
x->0

The < are less than or equal. I can't figure out how to put them in correctly.
This was on my final and it's driving me crazy! I couldn't decided if the limit was 0 or didn't exist. If you could point out my mistakes I'd appreciate it!
 
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No mistake. The limit is 0. -|sin(x)|<=g(x)*sin(x)<=|sin(x)|. The outer terms go to zero as z->0 so g(x)*sin(x) must also go to zero.
 

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