0112 - Math 31A 2010.01.12 MATH 31A DISCUSSION JED YANG 1....

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Unformatted text preview: Math 31A 2010.01.12 MATH 31A DISCUSSION JED YANG 1. Limits 1.1. Basic Limit Laws. Assume that lim x c f ( x ) and lim x c g ( x ) exist. Then: (a) Sum Law: lim x c ( f ( x ) + g ( x )) = lim x c f ( x ) + lim x c g ( x ) . (b) Constant Multiple Law: For any number k R , lim x c kf ( x ) = k lim x c f ( x ) . (c) Product Law: lim x c ( f ( x ) g ( x )) = parenleftBig lim x c f ( x ) parenrightBigparenleftBig lim x c g ( x ) parenrightBig . (d) Quotient Law: If lim x c g ( x ) negationslash = 0, then lim x c f ( x ) g ( x ) = lim x c f ( x ) lim x c g ( x ) . 1.2. Exercise 2.3.22. Evaluate the limit lim z 1 z- 1 + z z +1 . Solution. Recall that lim z 1 z = 1 and lim z 1 1 = 1. By the Quotient Law, lim z 1 z- 1 = lim z 1 1 lim z 1 z = 1 1 = 1. By the Sum Law, lim z 1 z- 1 + z = lim z 1 z- 1 + lim z 1 z = 1+1 = 2. By the Sum Law, lim z 1 z +1 = 2. So by the Quotient Law, lim z 1 z- 1 + z z +1 = lim z 1 z- 1 + z lim z 1 z +1 = 2 2 = 1. square 1.3. Exercise 2.3.29. Can the Quotient Law be applied to evaluate lim x sin x x ? Solution. The Quotient Law requires the limit of the denominator, namely, lim x x , to exist and be nonzero. This is not the case, so we cannot apply directly. square 1.4. Exercise 2.3.30. Show that the Product Law cannot be used to evaluate lim x / 2 ( x / 2) tan x . Solution. The Product Law requires the limit of each factor to exist. However, lim x / 2 tan x does not exist. square 1.5. Exercise 2.3.31. Give an example where lim x ( f ( x ) + g ( x )) exists but nei-...
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0112 - Math 31A 2010.01.12 MATH 31A DISCUSSION JED YANG 1....

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