12.8+OFSA+Solutions


 * OFSA SOLUTIONS for Limits**

This page contains peer generated solutions and error explanations to OFSA questions. As you read or view the solutions, be critical: check for accuracy, but also for more efficient solution strategies. If you have a better method or different idea/answer, post a discussion and monitor the responses.


 * Quick Directions**
 * Post answers, solutions and error explanations to each OFSA question below.
 * For each "distractor" or incorrect answer choice, explain the error that would lead to that incorrect answer choice.
 * You may either do the above in typed format or using a pencast.
 * Separate each question with a section bar.
 * After each solution, provide a hyperlink back to the corresponding OFSA page.
 * Follow example below.
 * Click here to refer to solution format in 7.7

=**Question 1**= toc Given math f(x)=3x^2-5x math evaluate math \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}h math

A. 6x+3h-5 B. 6x-5 C. 0 D. Does Not Exist (DNE)


 * __Solution 1__**

math \lim_{h\rightarrow 0}\frac{3(x+h)^2-5(x+h)-(3x^2-5x)}h math

math \lim_{h\rightarrow 0}\frac{3(x^2+2xh+h^2)-5x-5h-3x^2+5x}h math

math \lim_{h\rightarrow 0}\frac{3x^2+6xh+3h^2-5x-5h-3x^2+5x}h math

math \lim_{h\rightarrow 0}\frac{6xh+3h^2-5h}h math

math \lim_{h\rightarrow 0}\frac{6xh+3h^2-5h}h math

math \lim_{h\rightarrow 0}\frac{h(6x+3h-5)}h math

math \lim_{h\rightarrow 0}(6x+3h-5) math

math 6x-5 math


 * __Error Explanation 1__**

A. 6x+3h-5 //Error. Student forgot to evaluate the limit as h approaches the value 0 although they may have simplified the function to its simplest form (they forgot to find the limit using direct substitution.)//

B. 6x-5 //Correct!!!//

C. 0 //Error. Student only focused on manipulating the algebra in the numerator that they forgot to divide the entire numerator by h.//

D. Does Not Exist (DNE) //Error. Student forgot to distribute the negative when subtracting the entire f(x) function from the f(x+h) function.//

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=**Question 2**=

Evaluate math \lim_{n \to -\infty }\frac{x+9}{4x+5} math

A. 1/4 B. 1 C. 2/5 D. Indeterminant


 * __Solution 2__**

math \lim_{n \to -\infty }\frac{1+(9/n){}}{4+(5/n)} math

math \lim_{n \to -\infty }\frac{1+0}{4+0} math

math \frac{1}{4} math


 * __Error Explanation 2__**

A. 1/4 //Correct!!!//

B. 1 //Error. Student simply used direct substitution to solve the problem and got// math \frac{-\infty}{-\infty} math //Then, the student assumed that the result of this expression is 1. However, they do not realize that different levels of infinity exist.//

C. 2/5 //Error. Student thought// math "\frac{number}{\infty }" \rightarrow 1 math //when really// math "\frac{number}{\infty }" \rightarrow 0 math //for limits at infinity problems.//

D. Indeterminant //Error. Student simply used direct substitution to solve the problem and got// math \frac{-\infty}{-\infty} math //like in// B. //However, this student assumed that the result of this expression was indeterminant form and couldn't be taken any further.(So there could be two different interpretations of this expression.)//

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=**Question 3**=

Find the value of A that makes math \[f(x) = \begin{cases} \(1-2x) & \text{ if } x\leq -2 & (Ax+5) & \text{ if } x>-2 \end{cases}

\] math continuous.

A. f(a) does not exist, so therefore the function is not continuous B. 5 C. Undefined D. 0

__**Solution 3**__

math 1-2(-2)= A(-2)+5 math

math 1+4=-2A+5 math

math 0=-2A math

math A=0 math


 * __Error Explanation 3__**

A. f(a) does not exist, so therefore the function is not continuous //Error. Although the student demonstrates their understanding of the definition of continuity [f(a) must exist], they don't realize that f(-2) does, in fact, exist because the restrictions accommodate for the value of -2 from the left, at -2, and from the right.//

B. 5 //Error. Student only evaluated the first equation by substituting -2 in for x and thought this was the value of A that would make the function continuous.//

C. Undefined //Error. Student plugged in the value of -2 for A// //instead of x, in which case they got an expression// "1=5" //which is not true and would lead them to believe the answer is undefined.//

D. 0 //Correct!!!//

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=**Question 4**= Click here to return to OFSA.
 * __Solution 4__**
 * __Error Explanation 4__**

=**Question 5**= Click here to return to OFSA.
 * __Solution 5__**
 * __Error Explanation 5__**

=**Question 6**= Click here to return to OFSA.
 * __Solution 6__**
 * __Error Explanation 6__**

=**Question 7**= Click here to return to OFSA.
 * __Solution 7__**
 * __Error Explanation 7__**

=**Question 8**= Click here to return to OFSA.
 * __Solution 8__**
 * __Error Explanation 8__**

=**Question 9**= Click here to return to OFSA.
 * __Solution 9__**
 * __Error Explanation 9__**

=**Question 10**= Click here to return to OFSA.
 * __Solution 10__**
 * __Error Explanation 10__**

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 * Question 11**
 * __Solution 11__**
 * __Error Explanation 11__**

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 * Question 12**
 * __Solution 12__**
 * __Error Explanation 12__**

math \lim_{x\to \infty } \frac {x^{3}+4x-3}{x^{3}+2} math b. -3/2 c. DNE d. 2/3
 * Question 13**
 * __Solution 13__**
 * a. 1**

Step One: math \lim_{x\to \infty } \frac {(x^{3}+4x-3)/x^3}{(x^{3}+2)/x^3} math

Step 2 math \lim_{x\to \infty } \frac {1+\frac{4}{x^2}-\frac{3}{x^3}}{1+\frac{2}{x^3}} math

Remember, math \frac{1}{\infty} math is very to close 0, so we can count it as zero. IT CANNOT ACTUALLY BE ZERO

math \lim_{x\to \infty } \frac {1+ 0-0}{1+0} math

math \lim_{x\to \infty } 1 math

__Error Explanation 13__

b. -3/2 A student would get this answer if they were to substitute zero for x, forgetting to divide by the highest bottom exponent of X c. DNE A student would get this if they incorrectly assume that one cannot have a limit at infinity. d. 2/3 A student would get this if they substituted for 1 for x
 * a. 1 This is the correct answer :)**


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Question 14 math f(x)= \left\{\begin{matrix} ax^2+3\rightarrow x< 4 \\ \sqrt{x}+10\rightarrow x\geq 4 \end{matrix}\right. math

a.13 b.15/16 c. 9/16 d.Never can be continuous

__Solution 14__ Step 1: math ax^2+3=\sqrt{x}+10 math

Step 2: math a(4)^2+3=\sqrt{4}+10 math

Step 3: math 16a+3=12 math

Step 4: math a=9/16 math

Error Explanation 14 a.13- A student substitutes 0 in for X instead of 4. b. 15/16- If a student were rushing and were to add the 3 instead of subtract it, they would get this answer. c. 9/16 - correct answer d. Never Continuous- If a student incorrectly assumes that two equations will never meet, then it should be never continuous. This will never happen unless the all the lines given are parallel.


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Question 15 math \lim_{x\rightarrow 3} \frac{1}{x^2-9} math a.1/9 b. math \sqrt{\frac{28}{3}} math c.-1/9 d. DNE __Solution 15__ DOES NOT EXIST. Because it cannot be canceled out anywhere, it is asymptote that has 2 different Y-values as X approaches 3 from the positive and negative. __Error Explanation 15__ a. 1/9- Often, if a student has a DNE answer, they doubt themselves and will go looking for the best other option. This one is the opposite of answer C, so then they would look at this answer and begin to doubt what they know is true. This is a very blatant distractor. b. math \sqrt{\frac{28}{3}} math

- If the student were to be solving

math \frac{1}{x^2-9} = 3 math then they would get three as an answer. c.-1/9- A student would get this answer if they were to either substitute zero in, or if they were to disregard the X in total. d. DNE- Correct answer, see explanation above


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 * __ Question 16 __**

math Find: \lim_{x\rightarrow 0}\tfrac{\sqrt{x^{2}+9}-3}{x^{2}} math

a. DNE b. 1/3 c. 1/6 d. 1/12 e. none of the above

__Solution 16__ math Find: \lim_{x\rightarrow 0}\tfrac{\sqrt{x^{2}+9}-3}{x^{2}} math math (\tfrac{\sqrt{x^{2}+9}-3}{x^{2}})(\tfrac{\sqrt{x^{2}+9}+3}{\sqrt{x^{2}+9}+3}) math math \tfrac{x^{2}+9-9}{x^{2}(\sqrt{x^{2}+9}+3)} math math \tfrac{x^{2}}{x^{2}(\sqrt{x^{2}+9}+3)} math math \tfrac{1}{\sqrt{x^{2}+9}+3} math math \lim_{x\rightarrow 0}\tfrac{1}{\sqrt{x^{2}+9}+3}=\tfrac{1}{\sqrt{(0)^{2}+9}+3} math math =\tfrac{1}{\sqrt{9}+3}=\tfrac{1}{3+3}=\tfrac{1}{6} math math \lim_{x\rightarrow 0}\tfrac{\sqrt{x^{2}+9}-3}{x^{2}}=\tfrac{1}{6} math

__Error Explanation 16__ a. DNE- Student assumed that the limit doesn't exist because he only paid attention to the x in the denominator b. 1/3- Student was on the right track but, at the end, made a simple mistake by substituting the whole expression in the square instead of just x squared c. 1/6- Correct answer d. 1/12- Student was on the right track but, at the end, made a simple mistake by forgetting about the square root e. none of the above- Student did something else leading to an incorrect answer
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 * __ Question 17 __**

math Find: \lim_{x\rightarrow 0}\tfrac{\sqrt{2-x}-\sqrt{2}}{x} math

a. DNE b. math \frac{-\sqrt{2}}{2} math c. math \frac{-1}{\sqrt{2}} math d. math \frac{-\sqrt{2}}{4} math e. none of the above

__Solution 17__ math Find: \lim_{x\rightarrow 0}\tfrac{\sqrt{2-x}-\sqrt{2}}{x} math math \tfrac{\sqrt{2-x}-\sqrt{2}}{x}(\tfrac{\sqrt{2-x}+\sqrt{2}}{\sqrt{2-x}+\sqrt{2}}) math math \tfrac{-x}{x(\sqrt{2-x}+\sqrt{2})} math math \tfrac{-1}{(\sqrt{2-x}+\sqrt{2})} math math \lim_{x\rightarrow 0}\tfrac{\sqrt{2-x}-\sqrt{2}}{x}=\tfrac{-1}{(\sqrt{2-(0)}+\sqrt{2})} math math =\tfrac{-1}{(\sqrt{2}+\sqrt{2})}=\tfrac{-1}{(2\sqrt{2})}(\tfrac{\sqrt{2}}{\sqrt{2}})=\tfrac{-\sqrt{2}}{4} math math \lim_{x\rightarrow 0}\tfrac{\sqrt{2-x}-\sqrt{2}}{x}=\tfrac{-\sqrt{2}}{4} math

__Error Explanation 17__ a. DNE- Student assumed that the limit doesn't exist because he only paid attention to the x in the denominator b. math \frac{-\sqrt{2}}{2} math Student was close but saw only one square root of 2 when there were 2 square root of 2's c. math \frac{-1}{\sqrt{2}} math same as b because they are the same answer d. math \frac{-\sqrt{2}}{4} math correct answer e. none of the above- Student did something else leading to an incorrect answer


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 * __ Question 18 __**

math Find: \lim_{x\rightarrow 0}\left | x \right | math

a. DNE b. 1 c. 0 d. -1 e. none of the above

__Solution 18__ math Find: \lim_{x\rightarrow 0}\left | x \right | math math \lim_{x\rightarrow 0}\left | x \right |=\left | (0) \right |=0 math 0 is neither positive nor negative, so in this case, absolute value doesn't matter

__Error Explanation 18__ a. DNE- Student probably guessed b. 1- Student probably guessed c. 0- correct answer d. -1- Student probably guessed e. none of the above- Student did something else leading to an incorrect answer


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 * __ Question 19 __**

math Find: \lim_{x\rightarrow 0}\tfrac{ \left | x \right |}{x} math

a. DNE b. 1 c. 0 d. -1 e. none of the above

__Solution 19__ math Find: \lim_{x\rightarrow 0}\tfrac{ \left | x \right |}{x} math Finding the left and right side limits is the best thing to do in this problem math \lim_{x\rightarrow 0^{+}}\tfrac{ \left | x \right |}{x}=\lim_{x\rightarrow 0^{+}}\tfrac{x}{x}=\lim_{x\rightarrow 0^{+}}1=1 math math \lim_{x\rightarrow 0^{-}}\tfrac{ \left | x \right |}{x}=\lim_{x\rightarrow 0^{-}}\tfrac{-x}{x}=\lim_{x\rightarrow 0^{-}}-1=-1 math math \lim_{x\rightarrow 0}\tfrac{ \left | x \right |}{x}=DNE math

__Error Explanation 19__ a. DNE- correct answer b. 1- Student had the right idea but had mistaken the right side limit as the answer c. 0- Student either forgot about x in denominator or guessed d. -1- Student had the right idea but had mistaken the left side limit as the answer e. none of the above- Student did something else leading to an incorrect answer


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Question 20 __Solution 20__ __Error Explanation 20__
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Question 21 __Solution 21__ __Error Explanation 21__
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Question 22 __Solution 22__ __Error Explanation 22__
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Question 23 __Solution 23__ __Error Explanation 23__
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Question 24 __Solution 24__ __Error Explanation 24__** Click here to return to OFSA.

Student did something else leading to an incorrect answer