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 oddroot 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 onesided 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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 Spring '09
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 Calculus, Derivative, Limits, lim, Limit of a function

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