MATH 161
1.
Solutions: QUIZ IV
30th May 2017
[2 pts] Using an appropriate table (for example, letting x = ± 0.1, ±0.01, ±0.001, etc.) determine (to the
nearest hundredth) the limit of the function
f ( x)  (1  x)1 / x
as x → 0. (Show your work!)
(Note that this is another type of indeterminate form, 0∞ , that we have not yet explored.)
Solution:
x
f ( x)  (1  x)1 / x
0.1
2.59374
0.01
2.70481
10-3
2.71692
10-4
2.71815
10-5
2.71828
10-6
2.71828
10-7
2.71828
10-8
2.71828
-0.1
2.86797
-0.01
2.7320
-10-3
2.7864
-10-4
2.71830
-10-5
2.71828
-10-6
2.71828
Thus the answer to the nearest hundredth is 2.72
(Soon we will prove that, as x→0, f(x) → 2.71828…= e)
2. (a) [1 pt] Carefully state the Intermediate Value Theorem.
Let f be a continuous function on the interval [a, b]. Let L be any real number between f(a) and
f(b). Then there exists a real number p belonging to [a, b] such that f(p) = L.
2
(b) [1 pt] Using the IVT explain why the function f(x) = x + 3 – 2 sin x must have at least one real root.
Solution: Observe that G(0) = 3 > 0 and G(-) = - + 1 < 0. Note also that G is continuous on the interval
[-, ]. Since G(-) < 0 < G(0), the Intermediate Value Theorem guarantees that there is a number p between
0 and  /2 for which G(p) = 0.
2 x 3  x 2  3x
3. [2.5 pts] Consider the rational function F defined by F ( x) 
2 x 2  5 x  12
(a) Where is F undefined? (Hint: Your answer should consist of two x values.)
Solution: Begin by factoring:
2 x 3  x 2  3x x(2 x  3)( x  1)
F ( x)  2

2 x  5 x  12 (2 x  3)( x  4)
We see that F is undefined when x = - 3/2 or x = 4.
(b) Let p denote the smaller of the two numbers found in part (a). Is it possible to extend F to a function that
is continuous at x = p? If not, explain; if so, how should F(p) be defined?
Solution:
Note that p = -3/2. Then:
 3  5 
    
x(2 x  3)( x  1)
x( x  1)  2  2 
15
lim F ( x)  lim
 lim


3
11
(2 x  3)( x  4)
x4
22
x 

2
2
Hence F may be extended continuously to the point x = -3/2.
(c) Let q denote the larger of the two numbers found in part (a). Is it possible to extend F to a function that is
continuous at x = p? If not, explain; if so, how should F(p) be defined?
Solution:
Note that q = 4. Then:
x(2 x  3)( x  1)
x( x  1)
 lim
x  4 ( 2 x  3)( x  4)
x 4 x  4
lim F ( x)  lim
x 4
Since the denominator approaches 0 as x approaches 4, yet the numerator approaches 12, the
limit does not exist. Hence F cannot be extended continuously to the point x = 4.
3
4. [2 pts] Give the type of discontinuity for each function below at the given point.
(a) At x = -1
answer: infinite
(b) At x = 2
answer: removable
(c) At x = 2
answer: jump
(d) At x = 1
answer: essential
1.0
0.5
2
1
1
0.5
1.0
2
4