M160 Concept Quiz I



MATH 160 Concept Quiz Name: ______________________________

Date: ________________ MATH 160 section _____

Time limit: 20 minutes (5 min. to confer, 15 min to write) Calculator allowed.

1. a) Let f(x) = [pic] . Show algebraically that limx→1 f(x) = 2.

b) Let g(x) = [pic]. Since f(1) = 3 and g(1) = 2, f(x) and g(x) are certainly not the same function. This being the case, explain why you can find limx→1 f(x) by finding limx→1 g(x) instead.

2. Is limx→1 f(x) really equal to 2, or is 2 only an approximation for limx→1 f(x) ? Explain.

3. When we write limx→c f(x) = L we mean that for every ε > 0 there exists a corresponding δ > 0 such that for all x in the domain of f with 0 < |x – c| < δ we have |f(x) – L| < ε.

Sketch the relevant graph and use it to find a δ corresponding to ε = 0.2 in the limit statement in #1. In your sketch illustrate clearly how you found δ from the graph. (You may use your calculator.)

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4. In #3, you showed there exists a δ corresponding to ε = 0.2. Does the work you did (or should have done) in #3 show conclusively that all the conditions the definition requires in order that limx→1 [pic] = 2 are met? Explain why or why not.

Concept Quiz: Precise definition of limit

Scoring Notes

1. (a) (2 points) Let f(x) = [pic] . Show algebraically that limx→1 f(x) = 2.

+1 correct algebraic approach (rationalize denominator)

+1 for correct implementation and final evaluation

-1 for leaving out limit signs. (Judgment required when only some limit signs missing)

(No negative scores – i.e. incorrect approach (0 pts), incorrectly implemented (0 pts), no limit signs (-1 pt) does not yield a final score of -1.)

b) (2 points) Let g(x) = [pic]. Since f(1) = 3 and g(1) = 2, f(x) and g(x) are certainly not the same function. This being the case, explain why you can find limx→1 f(x) by finding limx→1 g(x) instead.

The limit depends on the values of the function f near x =1, not on its value at x = 1. Since f and g have the same values for values x near x =1 but ≠ 1 not on its value at x = 1 f and g have the same limit as x approaches 1.

2. (2 points) Is limx→1 f(x) really equal to 2, or is 2 only an approximation for limx→1 f(x) ? Explain.

The limit is really equal to 2. This is exactly and definitely the unique number that can be approximated arbitrarily accurately by evaluating the function using x values close enough to by not equal to 1.

+1 limx→1 f(x) is really equal to 2 +1 for reasonable explanation 0 pts for “only an approximation”.

3. (2 points) When we write limx→c f(x) = L we mean that for every ε > 0 there exists a corresponding δ > 0 such that for all x in the domain of f with 0 < |x – c| < δ we have |f(x) – L| < ε.

Sketch the relevant graph and use it to find a δ corresponding to ε = 0.2 in the limit statement in #1. In your sketch illustrate clearly how you found δ from the graph. (You may use your calculator.)

+1 Essentially correct graph of function

and relevant horizontal & vertical lines

+1 For clear indication he/she knows how

to get correct δ from figure. I saw no

examples where a student showed he/she

knew what was going on but did not get

the correct δ = .36. Some misread 0.2 as .02.

Forgive this misreading.

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4. (2 points) In #3, you showed there exists a δ corresponding to ε = 0.2. Does the work you did (or should have done) in #3 show conclusively that all the conditions the definition requires in order that limx→1 [pic] = 2 are met? Explain why or why not.

+1 No, does not show conclusively

+1 Must check all ε > 0; we only checked ε = .2.

NOTE: Students may also say “did not check all x’s”. I can’t really disagree, but I think most students would say that it’s obvious from the graph that all x’s within the illustrated δ of 1 yield function values within 0.2 of L = 2. To earn the point, however, they must mention the ε problem.

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