Expert: Scott A Wilson - 11/6/2009

Question
my problems (set 3):
2. What is the limit of (4-x²)/(x²-1) as x approaches ∞?
(A) 1, (B) 0, (C) -4, (D) -1, (E) ∞
4. What is the limit of (x/x) as x approaches 0?
(A) 1, (B) 0, (C) ∞, (D) -1, (E) nonexistent
6. What is the limit of (4-x²)/(4x²-x-2) as x approaches ∞?
(A) -2, (B) (-1/4), (C) 1, (D) 2, (E) nonexistent
7. Consider the function (this is a piecewise function) f(x)=(x²/x) if x≠0 or 3 if x=0. The function:
(A) is continuous everywhere
(B) is continuous except at x=0
(C) has a removable discontinuity at x=0
(D) has an infinite discontinuity at x=0
(E) has x=0 as a vertical asymptote
9. What is the limit of sin(x) as x approaches ∞?
(A) is nonexistent, (B) is infinity, (C) oscillates between -1 and 1 (D) is zero (E) is 1 or -1
10. Consider the function (this is a piecewise function) if f(x)=((x²-x)/(2x) for x≠0 or f(0)=k. If f is continuous at x=0, then what is k equal to?
(A) -1, (B) (-1/2), (C) 0, (D) (1/2), (E) 1
51. At which point(s) does the graph of the equation y=((1/3)(x³))-((3/2)(x²))+(2x) have a horizontal tangent line?
56. Find the value of a and b if the tangent to y=(a/x²)+b at (2,4) has slope m-sub-tan=-2.
58. Find k if the curve y=x²+k is tangent to the line y=2x.

Answer
There may be several question, but they won't take long to answer.

2 Divide numerator and denominator by x².  As x->∞, you get -1/1 = -1.

4 Limit x/x as x->0 is 1/1 = 1. A

6 Multiply numerator and denominator by 1/x² and the answer is -1/4.  B

7 The limit x-> f(x) is 0, but at f(0), the function is 3.
From this, the function is continuous everywhere but at 0.  B

9 sin(x) always oscillates between -1 and 1, so it has not limit. A

10 f(x) = (x²-x)/(2x) = x(x-1)/2x = (x-1)/2 = -1/2.  B

51 y' = x² - 3x + 2, so y' = (x-2)(x-1).
That means it has 0 slope at x=1 and x=2.

56 I don't understand the question.
The slope is y' = -2a/x³, and at x=2, the slope is -2a/8 = -a/4.

58 The slope of the line is 2, wand the slope of the parabola is 2x.
Since 2x=2 at x=1, the value of the line at x=1 is y = 2•1 => y = 2.
For (1,2) to be on the parabola, y = x² + k, we need 2 = 1² + k, which says k = 1.