document.write( "Question 1035838: Let $\"f%28x%29=%28x%5E2%28x-1%29%285x%5E2%2B3x-4%29%29%2F%282x%5E5-3x%5E4%2B4x%5E3-5x%5E2%29\"$
\n" ); document.write( "Find:
\n" ); document.write( "a. lim f(n) as n-->-Infinity
\n" ); document.write( "b. lim f(n) as n-->-1
\n" ); document.write( "c. lim f(n) as n-->0 \n" ); document.write( "

Algebra.Com's Answer #650516 by robertb(5830) $\"\"$ $\"About$

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a. Suppose $\"x+%3C%3E+0\"$ . Then $\"x%5E2\"$ can be cancelled from top and bottom, and the numerator is a polynomial in which the leading term is $\"5x%5E3\"$ , while the denominator is a polynomial with a leading term of $\"2x%5E3\"$ . Therefore as n goes to infinity, f(n) goes to 5/2.\r
\n" ); document.write( "\n" ); document.write( "b. As n approaches -1, f(n) approaches -2/7. (In fact f(x) is continuous at x = -1).\r
\n" ); document.write( "\n" ); document.write( "c. Returning to the argument used in (a), in any neighborhood of x = 0, f(x) is defined except at x = 0 itself. This means the limit, as n approaches 0, would be -4/5. (There is a removable discontinuity at x = 0.) \n" ); document.write( "

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