alder - Name: ______________________ Math 2253: Midterm...

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Unformatted text preview: Name: ______________________ Math 2253: Midterm Exam 1 Solutions 1. For the function f(x) graphed at the right, find the following limits or explain why they do not exist. =0 = -1 a. lim e. lim b. lim =0 f. lim =0 c. lim =0 g. lim = DNE 0 1 1 0 d. h. 2. Find the requested limits or explain why they do not exist. a. lim = by substitution b. lim lim 3 lim → +∞ from right & -∞ from left 1 find lim 3 ·1 √3 √3 3 i.e. limit does not exist (DNE) for 1 3. If we know that By the Sandwich thm: lim 3 2 lim thus lim 3 3 3 2 √3 3 √3 and √3 32 3 lim lim √3 lim . 4. Find the limits: a. lim c. lim d. lim b. lim lim 2 lim / 7 lim lim 7 Note that we can take an odd-root of any real. / so we have lim as 2 3 lim ∞ k · lim 2 =k =3 we see 2 becomes very small but remains negative, = –∞ i.e. unbounded in a negative direction / lim 7077 3 0 5. The function g(x) is defined on [-1, 3] by the graph on the right. a. Is g(x) cont on [-1, 3]? If not, where and why? Not continuous at x = 3, since limit ≠ functional value. b. Is g(x) diff’ble on [-1, 3]? If not, where and why? Not diff’ble @ x = 3 since not continuous there. Not diff’ble @ x = 2 since one-sided limits of diff quotient not equal. 6. At what points are the functions below continuous? a. 7 Everywhere except where denom. → 0, i.e. b. 2 | 3 1 Value in radical always positive, so continuous everywhere 7. Find the limit. Is the function continuous at t = 0? since tan (0) = 0; cos (0) = 1; sin (π/2) = 1 8. Define g (3) in a way that extends 2 2 lim sin cos tan 5 therefore continuous at x = 0 2 to be continuous on [-5, 5]. 6 3 would be continuous. 3 3 therefore Midterm 1 – Math 2253 Dr. Adler Page 1 Name: ______________________ 9. At what points does the graph of Giving the slope of curve = lim 2 2 2 5 have horizontal tangents? 2 2 Horiz tangent => m = 0 = 2 5, i.e. 2 5 2 5 i.e. , 2. 10. Using the limit of the difference quotient find an equation for the slope to the curve Then find the equation for tangent line to the curve at the point (-1, 1). and 2 2 2 2 2 2 2 Slope of curve = 1 lim 2 2 2 2 4 2 2 3 2 2 Equation of line through (-1, 1) with slope 2 is 2 y – 1 = 2 (x + 1) Midterm 1 – Math 2253 Dr. Adler Page 2 ...
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This note was uploaded on 07/05/2010 for the course MATH 2240 taught by Professor Crespo during the Spring '09 term at SPSU.

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