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document.write( "Factor the denominator using potential zeros and synthetic division.\r\n" );
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document.write( "Potential zeros are ±1, ±2. Try 1\r\n" );
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document.write( "1|1 0 -3 2\r\n" );
document.write( " | 1 1 -2\r\n" );
document.write( " 1 1 -2 0\r\n" );
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document.write( "So it works and thus we have factored the denominator as\r\n" );
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document.write( "(x-1)(x²+x-2)\r\n" );
document.write( "(x-1)(x-1)(x+2)\r\n" );
document.write( "(x-1)²(x+2)\r\n" );
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document.write( "The factor (x+2) does not become 0 when x=1, so if the\r\n" );
document.write( "numerator could be a multiple of (x-1)² then that factor\r\n" );
document.write( "would cancel into the top, creating a new function g(x)\r\n" );
document.write( "which is equal to f(x) for all x except 1. Its value\r\n" );
document.write( "at 1 would be defined, and thus f(x) would approach its\r\n" );
document.write( "value as x->1. So we will see if there is a constant k \r\n" );
document.write( "that would permit this cancellation. If so, we would\r\n" );
document.write( "have this identity:\r\n" );
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document.write( "
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document.write( "
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\r\n" );
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document.write( "So the constant terms would have to be the same:\r\n" );
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document.write( "k=-2. And if we substitute k=-2:\r\n" );
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document.write( "So they will be identical if a = -2. \r\n" );
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document.write( "So we substitute for the numerator in\r\n" );
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\r\n" );
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document.write( "and get\r\n" );
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document.write( "
\r\n" );
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document.write( "Now, after canceling, we have the new function \r\n" );
document.write( "g(x) which is identical to f(x) except for one\r\n" );
document.write( "point:\r\n" );
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document.write( "
\r\n" );
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document.write( "Then the limit of f(x) as x -> 1 will equal g(1) \r\n" );
document.write( "which is\r\n" );
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document.write( "
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document.write( "Answer: a=-2 and the limit = -2/3.\r\n" );
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document.write( "Edwin
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