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Math 265 Fall 2014 Exam 1 Review KEY draft
1. ! ! !!(!)
!!! = !
!!!
2.
a. Average velocity from t=2 to t=3: -‐50 ft/sec b. Average velocity from t=2 to t=2.5: -‐42 ft/sec c. Average velocity from t=2 to t=2+h: (-‐34 – 16h) ft/sec d. Instantaneous velocity at t=2, using c) above: -‐34 ft/sec as h goes to 0
3. 𝑓 𝑟 = 2𝜋𝑟! + !"##!
4. 𝑉 𝑥 = 300𝑥 − !!
!
5.
a. lim!→ !!!!!!!!!!"!!!!"
= !"!
b. lim!→ !"
!!!"! !!
= 10
c. lim!→ !!!!!!!!!
= −1
d. lim!→ !!!! !!! !
= ∞
e. lim!→!!!!!!!!!!!
= −∞
f. lim!→ !!!!!
!!(!!!!)= −∞
6. 𝑓 𝑥 =!!!!!!!!
if 𝑥 ≠ ±10 if 𝑥 = ±1
a. f is discontinuous at 𝑥 = 1 because lim!→! 𝑓(𝑥) does not exist since
lim!→!! 𝑓(𝑥) = −∞ whereas lim!→!! 𝑓(𝑥) = ∞ f is discontinuous at 𝑥 = −1 because lim!→!! 𝑓(𝑥) = − !
!≠ 𝑓 −1 = 0
b. Discontinuity at 𝑥 = 1 is infinite, while discontinuity at 𝑥 = −1 is removable.
7. a. Evaluate the following limits:
i. lim!→!!! 𝑓(𝑥) = 1 ii. lim!→!!! 𝑓(𝑥) = ∞ iii. lim!→!! 𝑓 𝑥 DNE iv. lim!→!! 𝑓(𝑥) = 2 v. lim!→!! 𝑓 𝑥 = 2 vi. lim!→! 𝑓 𝑥 = 2
Math 265 Fall 2014 Exam 1 Review KEY draft
b. Intervals where the function is continuous: (−∞,−1) ∪ (−1,2) ∪ (2,∞) c. Function is discontinuous at -‐1 because lim!→!! 𝑓 𝑥 DNE
with infinite discontinuity d. Function is discontinuous at 2 because lim!→! 𝑓 𝑥 = 2 ≠ 𝑓 2 = −1. This
discontinuity is removable.
8. Equations of the vertical asymptotes: x=-‐2. 𝑓 𝑥 = !!!!!!!!!!!!!!
= !!!!!!!
lim!→!!! 𝑓(𝑥) = ∞; lim!→!!! 𝑓(𝑥) = −∞
9. Since 𝑓 0 = −4 𝑎𝑛𝑑 𝑓 1 = 1, then by IVT, there exists a 𝑐 𝑖𝑛 (0,1) such that
𝑓 𝑐 = 0, 𝑡ℎ𝑎𝑡 𝑖𝑠 2𝑐! + 3𝑐 − 4 = 0.
10. Use Squeeze Theorem to show that lim! → ! 𝑥! sin!!= 0.
Since −1 ≤ sin !!≤ 1, we can multiply all sides by 𝑥! which is always nonnegative,
obtaining −𝑥! ≤ 𝑥!sin !!≤ 𝑥!. Now lim!→! −𝑥! = 0 and lim!→! 𝑥! = 0, so therefore
by Squeeze Theorem, lim! → ! 𝑥! sin!!= 0
11. If 4x + 5 ≤ f(x) ≤ x2 + x -‐ 5, we have lim!→!! 4𝑥 + 5 = −3, and
lim!→!! 𝑥! + 𝑥 − 5 = −3. Therefore by Squeeze Theorem lim!→!! 𝑓 𝑥 = −3
12. For the limit: lim!→! 𝑥! − 10 = 6, illustrate the definition by finding the largest possible value of δ that correspond to ε = 0.2.
a. We need to find δ>0 such that if 0 < 𝑥 − 4 < 𝛿, then 𝑥! − 10 − 6 < 0.2. But 𝑥! − 10 − 6 < 0.2 means that −.2 < 𝑥! − 16 < .2. Adding 16 to all sides, we
get 15.8 < 𝑥! < 16.2; taking square roots, we get 3.9749 < 𝑥 < 4.0249; Subtracting 4 from all side, we get −.0251 < 𝑥 − 4 < .0249. So we choose 𝛿 = .0249.
b. Proof: (Optional) Choose 𝛿 = .0249 If 0 < 𝑥 − 4 < .0249, then −.0249 < 𝑥 − 4 < .0249; which implies that 3.9751 < 𝑥 < 4.0249; which further implies that 15.8014 < 𝑥! < 16.1998; which means that −.1986 < 𝑥! − 16 < .1998; This means that 𝑥! − 16 < .1998 < .2; Therefore lim!→! 𝑥! − 10 = 6
13. Prove that lim! →! 2𝑥 − 4 = 2 using 𝜀 − 𝛿 definition.
14. Sketch a graph of a function f defined on [-‐2, -‐1) ∪ (−1, 3] and continuous on
[-‐2, -‐1) ∪ (-‐1, 3) that satisfies given conditions: 𝑓 −2 = 0; 𝑓 1 = 0; 𝑓 3 = −2; lim
! → !!!𝑓 𝑥 = ∞; lim
! → !!!𝑓 𝑥 = −∞; lim
! → !!𝑓 𝑥 = 2;
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