document.write( "Question 988009: Use the definition of limit to verify that:
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Algebra.Com's Answer #608667 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "We are to prove that no matter how small a tolerance e > 0 we desire for\r\n" );
document.write( "(5-2x) to be within -1, we will always be able to find at least one value \r\n" );
document.write( "for d > 0 such that whenever x is within d of 3, that (5-2x) will always be \r\n" );
document.write( "within e of -1.  \r\n" );
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document.write( "Suppose we are given e > 0 and desire to prove that there exists a value of \r\n" );
document.write( "d > 0 such that |f(x)-(-1)| < e.  We look at the absolute value of the \r\n" );
document.write( "difference between the (5-2x) and -1\r\n" );
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document.write( "|(5-2x)-(-1)| = |5-2x+1| = |6-2x| = 2|3-x| = 2|x-3| we want to be < e\r\n" );
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document.write( "So the (5-2x) will be within e of -3 if 2|x-3| < e, which is to say\r\n" );
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document.write( "when |x-3| < e/2, so if we choose any d < e/2, then when |x-3| < d, that is,\r\n" );
document.write( "when x is within d of 3, (5-2x) is within e of 3.                                    \r\n" );
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document.write( "So given e > 0 |(5-2x)-(-1)| < e whenever d < e/2.\r\n" );
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document.write( "That proves that \r\n" );
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document.write( "Edwin

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