(Section 2.3: Limits and Infinity I) 2.3.19 Example 14Evaluate limx±²fx(), where =x2±3x3+4x2+1. Solution 1 (A Rigorous Solution): Left to the reader! Refer to Example 13. Solution 2 (The “Short Cut”: Dominant Term Substitution)limx=limxx2³3x3+4x2+1Indeterminate Limit Form ²²´µ¶·¸¹=limxx2x3=limx1xLimit Form 1²´µ¶·¸¹=0“Super Short Cut”: This is because the numerator, x2±3, has a degree (i.e., 2) that is less than the degree (i.e., 3) of the denominator,
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This note was uploaded on 12/29/2011 for the course MATH 150 taught by Professor Sturst during the Fall '10 term at SUNY Stony Brook.