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From 1-8 chose the correct answer

From 9-17 answer explanation

 Suppose that imit as (x to infinity) f(x) = infinityand imit as (x to infinity) g(x) =-infinity. Then always,

(A) imit as (x to infinity) f(x)*g(x) = infinity

(B) imit as (x to infinity) f(x)*g(x) =-infinity

(C) imit as (x to infinity) f(x)+g(x) = infinity

(D) imit as (x to infinity) [f(x)]/[g(x)] = infinity

 Suppose that imit as (x to infinity) f(x) = infinityand imit as (x to infinity) g(x) = 0. Then always,

(A) imit as (x to infinity) f(x)*g(x) = infinity

(B) imit as (x to infinity) f(x)*g(x) = 0

(C) imit as (x to infinity) f(x)-g(x) = infinity

(D) imit as (x to infinity) [f(x)]/[g(x)] = infinity.

 Suppose that imit as (x to a) f(x) = infinity. Then,

(A) imit as (x to a) 1/[f(x)] = infinity

(B) imit as (x to a) 1/[f(x)] = 0

(C) imit as (x to a) 1/[f(x)] =-infinity

(D) none of the above.

 Suppose that imit as (x to infinity) f(x) = 0but (x) > 0. Then,

(A) imit as (x to infinity) 1/[f(x)] = infinity

(B) imit as (x to infinity) 1/[f(x)] =-infinity

(C) imit as (x to infinity) 1/[f(x)] = 0

(D) imit as (x to infinity) 1/[f(x)] < 0.

 The function (x)assumes only two values: http://calculus.sfsu.edu/latexrender?src=%5Cbegin%7Bdisplaymath%7D+%5Ctextstyle%7B5%7D+%5Cend%7Bdisplaymath%7D&type=png&size=15and http://calculus.sfsu.edu/latexrender?src=%5Cbegin%7Bdisplaymath%7D+%5Ctextstyle%7B7%7D+%5Cend%7Bdisplaymath%7D&type=png&size=15. No matter how big http://calculus.sfsu.edu/latexrender?src=%5Cbegin%7Bdisplaymath%7D+%5Ctextstyle%7BN%7D+%5Cend%7Bdisplaymath%7D&type=png&size=15is, there is some > Nsuch that (x) = 5and there is some > Nsuch that (x) = 7. We conclude that

(A) imit as (x to infinity) f(x) = 5

(B) imit as (x to infinity) f(x) = 7

(C) imit as (x to infinity) f(x) = 6

(D) imit as (x to infinity) f(x)does not exist.

 Let (x) = 1/[x^2]. In order to ensure that (x) > N, it is enough to require that

(A) > N

(B) < N

(C) < 1/[root(N)]

(D) x| < 1/[root(N)].

 Which of the statements is true?

(A) imit as (t to infinity) sin t = 1

(B) imit as (t to infinity) sin t = pi

(C) imit as (t to infinity) sin t = infinity

(D) imit as (t to infinity) sin tdoes not exist.

 Which of the statements is true?

(A) imit as (t to infinity) cos 1/t = 1

(B) imit as (t to infinity) cos 1/t = pi

(C) imit as (t to infinity) cos 1/t = infinity

(D) imit as (t to infinity) cos 1/tdoes not exist.

 Graph the appropriate functions (first try by hand and then verify your sketches with the grapher) and then determine the infinite limits. Include small sketches. (a) imit as (x to 0^+) ln x,     (b) imit as (x to 0^+) 1/x,     (c) imit as (x to 0^-) 1/x,     (d) imit as (x to 0) 1/x,     (e) imit as (x to 0^+) 1/[x^2],    (f) imit as (x to 0^-) 1/[x^2],     (g) imit as (x to 0^-) cot x,     (h) imit as (x to 0^+) cot x,    (i) imit as (x to 0) cot x.   

 Use algebra and limits laws to find the limits:

(a) imit as (x to infinity) [x+1]/[x-2],

(b) imit as (x to-infinity) [x^2+1]/[x^2-5]

(c) imit as (x to infinity) [x^2+1]/[x^3-2]

(d) imit as (x to-infinity) [x^3+1]/[x^2+2].

Check your answers by graphing the functions.

 Use algebra and limits laws to find the limits: (a) imit as (x to infinity) (root (x+1)-root(x)),    (b) imit as (x to infinity) (x^3-x^2). Check your answers by graphing the functions, for example: graph = root (x+1)-root(x)for leq x leq 10000. You may leave the http://calculus.sfsu.edu/latexrender?src=%5Cbegin%7Bdisplaymath%7D+%5Ctextstyle%7By%7D+%5Cend%7Bdisplaymath%7D&type=png&size=15-range blank.

 Find the horizontal and vertical asymptotes of (x) = [x^2-3]/[x^2-1].

 Find the horizontal and vertical asymptotes of (x) = [x+1]/[x^2-5].

 Find the limit imit as (x to infinity) root ([12 x^3-4 x+1]/[1+3 x+2 x^3]).

 Let Does ttp://calculus.sfsu.edu/latexrender/pictures/3532ffa24f64b217f9c5cc798527fe67.png does ttp://calculus.sfsu.edu/latexrender/pictures/6d1f1add7c9fcb7b633d67c2dc8b184d.pnghave a limit as to infinity? Explain your answer.

 Let Does ttp://calculus.sfsu.edu/latexrender/pictures/3c3289fb5e863ecb2ce878f9d3756d27.pngdoes ttp://calculus.sfsu.edu/latexrender/pictures/6d1f1add7c9fcb7b633d67c2dc8b184d.pnghave a limit as to infinity? Explain your answer.

 The function (x) = 1/x+3has a limit = 3as to infinity. This means that if http://calculus.sfsu.edu/latexrender?src=%5Cbegin%7Bdisplaymath%7D+%5Ctextstyle%7Bx%7D+%5Cend%7Bdisplaymath%7D&type=png&size=15is sufficiently large (that is if > Nfor some number http://calculus.sfsu.edu/latexrender?src=%5Cbegin%7Bdisplaymath%7D+%5Ctextstyle%7BN%7D+%5Cend%7Bdisplaymath%7D&type=png&size=15), the values of (x)are closer to = 3than a number psilon > 0. (a) Sketch the graph = 1/x+3and a horizontal strip such that .8 < y < 3.2(if you are using the grapher, note that it has the capability to display more than one graph at once. The functions to be plotted are separated by a semincolon; you can input 1/x+3; 2.8; 3.2 for this problem). Based on the graph decide which portion of the graph = f(x)is contained in the strip. That is find http://calculus.sfsu.edu/latexrender?src=%5Cbegin%7Bdisplaymath%7D+%5Ctextstyle%7BN%7D+%5Cend%7Bdisplaymath%7D&type=png&size=15such that for > Nthe graph is included in the strip.

(b) For the following values of psilon, using algebraic inequalities find http://calculus.sfsu.edu/latexrender?src=%5Cbegin%7Bdisplaymath%7D+%5Ctextstyle%7BN%7D+%5Cend%7Bdisplaymath%7D&type=png&size=15such that the graph = 1/x+3restricted to > Nis contained in the horizontal strip -epsilon < y < 3+epsilon? Note that the first part of this inequality is always satisfied so it is enough to solve

< 3+epsilon

for http://calculus.sfsu.edu/latexrender?src=%5Cbegin%7Bdisplaymath%7D+%5Ctextstyle%7Bx%7D+%5Cend%7Bdisplaymath%7D&type=png&size=15.

ttp://calculus.sfsu.edu/latexrender/pictures/7c02d2643f97460ef94afde8f3803614.png

(c) What is a formula for http://calculus.sfsu.edu/latexrender?src=%5Cbegin%7Bdisplaymath%7D+%5Ctextstyle%7BN%7D+%5Cend%7Bdisplaymath%7D&type=png&size=15? (in terms of psilon).