document.write( "Question 1163650: Prove lim x^3 = 1 as x approaches 1 using precise definition of limits\r
\n" ); document.write( "\n" ); document.write( "This is how far I got and my line of thought not sure if I am making any conceptual errors.\r
\n" ); document.write( "\n" ); document.write( "Basically the question wants me to show that whenever they give me a e>0 value, I can return a &>0 value whereby
\n" ); document.write( "0<|x-1|<& --> |x^3 - 1|\n" ); document.write( "\n" ); document.write( "I tried to express the right hand inequality in terms of |x-1| so that I can relate to the left hand inequality.
\n" ); document.write( "|(x-1)(x^2 + x + 1)|\n" ); document.write( "|(x-1)((x-1)^2 + 3(x-1) - 3)|\n" ); document.write( "If I were to modulus all the x-1, I will get this inequality
\n" ); document.write( "||x-1|(|x-1|^2 + 3|x-1| - 3)| >= |(x-1)((x-1)^2 + 3(x-1) - 3)|
\n" ); document.write( "And if I were to convert it further to & terms,
\n" ); document.write( "&(&^2 + 3& - 3) > ||x-1|(|x-1|^2 + 3|x-1| - 3)| < e
\n" ); document.write( "&(&^2 + 3& -3) < e
\n" ); document.write( "So from here I am not too sure how to go about it. Please advise \n" ); document.write( "

Algebra.Com's Answer #787813 by Edwin McCravy(20055) $\"\"$ $\"About$

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( "|x³ - 1| < ε whenever |x - 1| < δ \r\n" );
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document.write( "|x³ - 1| < ε iff\r\n" );
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document.write( "|(x - 1)(x² + x + 1)| < ε iff\r\n" );
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document.write( "To find the appropriate δ on the interval (0,2), we know that\r\n" );
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document.write( "|x² + x + 1| < 1  (the value of this increasing function when x=0)\r\n" );
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document.write( "So on the interval (0,2), \r\n" );
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document.write( "|(x - 1)(x² + x + 1)| < ε iff \r\n" );
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document.write( "Thus whenever δ < ε, then \r\n" );
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document.write( "|x³ - 1| < ε \r\n" );
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document.write( "thus      [PROVED]\r\n" );
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document.write( "Edwin

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