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FinalSp04soln

# FinalSp04soln - MATH 118 SAMPLE FINAL SOLUTIONS FINAL EXAM...

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MATH 118 SAMPLE FINAL SOLUTIONS FINAL EXAM GIVEN MAY 4, 2004 Problem 1. Let f ( x ) = 1 x + 1 , -∞ < x < - 1 1 + x 1 - x , - 1 x < 1 2 x 2 - 1 x 2 + 1 1 x < +∞ a) (5 points) List all the values of x for which f ( x ) is not continuous. Calculate the following limits. If the limit is infinite, indicate whether it is +∞ or -∞ . b) (3 points) lim x 1 + f ( x ) c) (3 points) lim x 1 - f ( x ) d) (3 points) lim x →- 1 + f ( x ) e) (3 points) lim x →- 1 - f ( x ) f) (3 points) lim x →+∞ f ( x ) g) (3 points) lim x →-∞ f ( x ) Solution. a) You really need to complete the rest of the parts of this problem be- fore you do this part. Go ahead and do that now, then come back and read this solution. OK, you’re back? The function is not continuous at x = - 1 and at x = 1. This is because the limits from the left and from the right aren’t equal at these two points. (Part b) and c): 1 / 2 = +∞ ; Part d) and e): 0 = -∞ .) At every other point, f ( x ) is a rational function nearby x , whose denominator is never 0, and is therefore continuous at x . b) As x 1 + , i.e. as x approaches 1 from the right, the values of x are ( 2 x 2 - 1 )/( x 2 + 1 ) , and by the limit theorems this approaches ( 2 - 1 )/( 1 + 1 ) = 1 / 2: lim x 1 + f ( x ) = 1 2 . c) As x 1 - , i.e. as x approaches 1 from the left, the value of f ( x ) will be ( 1 + x )/( 1 - x ) as soon as x is close enough to 1. The numerator of this approaches 2 and the denominator approaches 0; this increases without bound in absolute value. Since the quotient is always positive for x in this range, therefore lim x 1 - f ( x ) = +∞ . d) As x → - 1 + , i.e. as x approaches - 1 from the right, the values of f ( x ) are governed by the middle condition, f ( x ) = 1 + x 1 - x . The numerator approaches 0 and the denominator approaches 2, there- fore by the limit theorems the quotient approaches 0: lim x →- 1 + f ( x ) = 0 . e) As x → - 1 - , i.e. as x approaches - 1 from the left, the values of f ( x ) are govered by the first condition, f ( x ) = 1 1 + x . The numerator is constantly 1 and the denominator approaches 0, so the quotient “blows up” (becomes infinite in absolute value). Since the quotient is always negative for x < - 1, therefore lim x 1 - f ( x ) = -∞ . f) As x → +∞ , the value of f ( x ) is ( 2 x 2 - 1 )/( x 2 + 1 ) . We rewrite this by dividing numerator and denominator by x 2 , f ( x ) = 2 - 1 x 2 1 + 1 x 2 . In this form it’s clear that 1 / x 2 0, and thus lim x →∞ f ( x ) = 2 1 = 2 .

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SOLUTIONS g) As x → -∞ , the value of f ( x ) is 1 /( 1 - x ) , and therefore lim x →-∞ f ( x ) = 0 . (Divide the numerator and denominator by x and use the fact that 1 / x 0 as x → -∞ .) Problem 2. Find an equation of the tangent line to the curve x 3 - 3 y 3 + 2 x + y = 1 at the point x = 1 , y = 1 . Solution. To find the tangent line we need to find dy / dx , which we do by implicit differentiation: d dx x 3 - 3 y 3 + 2 x + y = d dx 1 , i.e. 3 x 2 - 9 y 2 dy dx + 2 + dy dx = 0 which we solve for dy / dx : dy dx = - 3 x 2 - 2 - 9 y 2 + 1 .
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