document.write( "Question 1163529: Prove lim (2x^2 - x - 1) = 2 as x approaches -1 using precise definition of limits.\r
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Algebra.Com's Answer #787712 by Edwin McCravy(20054) $\"\"$ $\"About$

You can put this solution on YOUR website!

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document.write( "For any given ε > 0, we must find a δ > 0 such that\r\n" );
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document.write( "|(2x² - x - 1) - 2| < ε whenever |x - (-1)|  = |x + 1| < δ \r\n" );
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document.write( "|(2x² - x - 1) - 2| < ε iff\r\n" );
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document.write( "|2x² - x - 1 - 2| < ε iff\r\n" );
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document.write( "|2x² - x - 3| < ε iff\r\n" );
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document.write( "|(2x-3)(x+1)| < ε\r\n" );
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document.write( "To find the appropriate δ on the interval (-2,0), we know that\r\n" );
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document.write( "|2x-3| < 7\r\n" );
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document.write( "So on the interval (-2,0), \r\n" );
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document.write( "|(2x-3)(x+1)| < ε iff \r\n" );
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document.write( "Thus whenever δ < ε/7, then \r\n" );
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document.write( "|(2x² - x - 1) - 2| < ε \r\n" );
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document.write( "thus      [PROVED]\r\n" );
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document.write( "Edwin

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