Calculus I Practice Midterm 1 Solutions
                Instructions
                ❼ Write your name and UNI clearly in the section below.
                ❼ You are NOT allowed to use class notes, books and homework solutions in the exam-
                  ination.
                ❼ Except for True/False questions, show all computations and work in your answer.
                ❼ Don’t cheat! If it looks like you are cheating, then you are cheating.
                                           Question  Points  Score
                                               1       10
                                               2       10
                                               3        4
                                               4        6
                                               5       10
                                               6        5
                                               7        5
                                            Total:     50
             Name:
             UNI:
                                                 Page 1 of 8
              Math 1101                            Calculus I          Practice Midterm 1 Solutions
               1. (10 points) True/False 2 points each
                  (a) T    F f(x)=sin(x2) is an even function.
                  (b) T    F Thegraph of f(2x) is obtained from stretching the graph of f(x)
                               horizontally by a factor of 2.
                  (c) T    F Wehave that
                                            limxsin1= limx·limsin1
                                           x→0       x     x→0   x→0      x
                  (d) T    F Thefunction f(x) = x6 +x−1 has a solution in (0,1).
                  (e) T    F Thederivative of 1 is 1.
                 (You may use this area as scratchwork.)
                   Solution:
                                                            2         2
                   (a) T. We compute that f(−x) = sin((−x) ) = sin(x ) = f(x). Therefore f(x) =
                       sin(x) is even.
                   (b) F. The graph of f(2x) is obtained from shrinking the graph of f(x) horizontally
                       by a factor of 2.
                   (c) F. Because                   
                                            limsin   1   DOESNOTEXIST
                                            x→0      x
                       we cannot use the Product limit law and thus the equation above is false.
                   (d) T. We compute that f(0) = −1 and f(1) = 1. Because polynomials are contin-
                       uous at all real numbers and in particular in the interval [0,1] the Intermediate
                       ValueTheoremshowsthatf(x)mustequal0atsomepointin(0,1)andtherefore
                       f(x) has a solution in (0,1).
                   (e) FThederivativeof1iszero, eitherbyanexplicitcomputationusingthedefintion
                                                       0                               ′     0
                       of the derivative, or noting that x = 1 and so by the power rule (1) = (x ) =
                       0x−1 = 0.
                                                   Page 2 of 8
               Math 1101                                 Calculus I            Practice Midterm 1 Solutions
                2. Compute the following limits, if they exist. If the limit does not exist, explain why.
                    (a) (3 points) lim     x−2
                                    x→3 2
                                        x −5x+6
                          Solution:
                                               x−2                  x−2        x6=3       1
                                        lim 2            =lim                  ==lim
                                        x→3 x −5x+6        x→3 (x −2)(x−3)         x→3 (x −3)
                          Note     1     goes to infinity at x = 3 and thus the limit does not exist. To
                                (x−3)
                          be more precise, we will show that the right and left handed limits are not the
                          same.                1               1               1             1
                                         lim        =Rlim          =lim               =lim
                                           +            x→3          h→0                 h→0
                                        x→3 x−3              x−3 h>0(3+h)−3              h>0 h
                          Because h > 0, the quantity above is always positive. If we repeat the same
                          calculation with the left handed limit however, we find
                                        lim    1   =Llim 1 =lim               1      =lim 1
                                          −            x→3          h→0                 h→0
                                       x→3 x−3              x−3 h>0(3−h)−3              h>0 −h
                          Because h > 0, the quantity above is always negative. Since a positive number
                          is never equal to a negative number we conclude that
                                                       lim    1   6= lim    1
                                                          −             +
                                                      x→3 x−3        x→3 x−3
                          and therefore the limit doesn’t exist.
                    (b) (3 points) limx4sin1
                                    x→0         x
                          Solution: Notice that −1 ≤ sin1 ≤ 1. Because x4 ≥ 0 for any value of x,
                                                              x
                          it follows that we have the inequality
                                                        −x4 ≤sin1≤x4
                                                                     x
                          Notice that lim−x4 = limx4 = 0 because polynomials are continuous so we
                                       x→0         x→0
                          can just plug in 0 to evaluate the limit. Therefore by the Squeeze Theorem it
                          follows that                              
                                                         limx4sin    1   =0
                                                         x→0         x
                                                         Page 3 of 8
               Math 1101                                Calculus I           Practice Midterm 1 Solutions
                                          √          √       
                    (c) (4 points) lim cos    2+x− 2−x
                                   x→0              x
                          Solution: Because cos(x) is continuous at all real numbers, we can bring the
                          limit inside, e.g.
                                          √           √                  √          √      
                                   limcos     2+x− 2−x =cos lim 2+x− 2−x                              (1)
                                   x→0               x                  x→0         x
                          Wenowcompute the limit inside by rationalizing the numerator.
                              √         √              √         √        √         √
                                2+x− 2−x                 2+x− 2−x           2+x+ 2−x
                          lim                   =lim                     · √        √
                          x→0         x            x→0 √       x    √       2+x+ 2−x
                                                                2            2
                                                       ( 2+x) −( 2−x)                  (2 +x)−(2−x)
                                                =lim      √         √         =lim     √         √
                                                   x→0 x( 2+x+ 2−x)              x→0 x( 2+x+ 2−x)
                                                         √      2x √        x6=0     √       2 √
                                                =lim                        ==lim
                                                   x→0 x( 2+x+ 2−x)             x→0 ( 2+x+ 2−x)
                          Now notice that function in the final expression above is continuous at x = 0
                          because the denominator is not 0. Therefore by continuity we can plug in 0 to
                          evaluate the limit and find that
                                             √         √
                                               2+x− 2−x               2         2       1
                                         lim                   =√ √ = √ =√
                                         x→0         x              2+ 2       2 2       2
                          To obtain the final answer we plug this back into Equation (1) and find
                                                    √          √                  
                                                        2+x− 2−x                   1
                                            limcos                        =cos √
                                            x→0               x                     2
                3. Please give formal definitions below.
                    (a) (2 points) What does it mean for a function f(x) to be continuous at a point a?
                          Solution: f(x) is continuous at a point a if the both conditions are satisfied
                             ❼ limf(x) exists
                               x→a
                             ❼ limf(x) = f(a) (f(x) has the Direct Substitution Property at a.)
                               x→a
                   (b) (2 points) What does it mean for a function f(x) to be differentiable at a point a?
                          Solution: f(x) is differentiable at the point a if
                                           lim f(x)−f(a) = lim f(a+h)−f(a) exists
                                           x→a    x−a         h→0        h
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