36. Find the limits. lim x 
e x  e x
e x  e x
Solution:
By multiplying numerator and denominator by e x , we get
e x  e x e x
lim x  x  x  x
e e e
e x  e x  e x  e x
 lim x  x x  x x
e e  e e
e x  x  e x  x
e 2 x  e0
e2 x  1
 lim x  x  x  x  x  lim x  2 x 0  lim x  2 x
e e
e e
e 1
Using theorem

1.2.2 theorem :
Let a be a real number , and suppose that
limxa f  x   L1
and
limx a g  x   L2
That is, the limits exist and have values L1 and L2 , respectively. Then :
 a  limxa [ f  x   g  x ]  limxa f  x   limxa g  x   L1  L2
lim x    e2 x   lim x  1
lim x    e2 x   lim x  1
Using the fact lim x  e2 x  0

0 1
 1
0 1
Answer:
lim x 
e x  e x
 1
e x  e x
 2 
38. Find the limits. lim x 0 ln  2 
x 
Solution:
2 
 2 

lim x 0 ln  2   ln  lim x 0 2   ln     
x 
x 

39. Find the limits. lim x  
 x  1
x
xx
Solution:
lim x 
 x  1
xx
x
 x 1 
 x 1
 1
 lim x  
  lim x      lim x  1    e
 x 
 x x
 x
x
x
x
Example 3: Figure 1.3.4 is the graph of f(x) = (1 + 1/x)x . As suggested by this graph
Figure 1.3.4
so the line y = e is a horizontal asymptote for f in both the positive and negative directions
21–26 Sketch a possible graph for a function f with the specified properties. (Many different solutions are
possible.) ■
21.
(i) the domain of f is [−1, 1]
(ii) f (1)  f (0)  f (1)  0
(iii) limx 1 f  x   limx 0 f  x   limx 1 f  x   1
Solution:
The graph is plotted at every point on the required domain and nowhere else. The value of the function is
zero at x=−1, x=0, and x=1.
The limits are also correct as specified.
22.
(i) the domain of f is [−2, 1]
(ii) f (2)  f (0)  f (1)  0
(iii) limx 2 f  x   2, limx 0 f  x   0, and limx1 f  x   1
This is one possible graph but answers may vary. Since the limits of f(x) as x approaches -2 and 1 do not
match up with the function values at those points, holes are necessary in order to plot lone points that are
separate from the curve, as shown in the graph to the left.
https://quizlet.com/explanations/textbook-solutions/calculus-early-transcendentals-10th-edition9780470647691?funnelUUID=00369c63-1780-4e7d-9576-eb51e3a0ffa2
23.
(i) the domain of f is (−ꝏ, 0]
(ii) f (2)  f (0)  1
(iii) limx 2 f  x    
This is just one of many possible solutions, your answer may vary. Take note of the necessity of having a
lone point at (-2, 1) since the limit of the function as x approaches -2 from BOTH sides is positive infinity,
which is why the curve increases without bound as shown at x = -2.
24.
(i) the domain of f is (0, + ꝏ)
(ii) f (1)  0
(iii) the y-axis is a vertical asymptote for the graph of f
(iv) f(x) < 0 if 0 < x < 1
This is just one possible solution, and your answer may vary. Our domain is clearly from 0 to positive
infinity as our curve goes to the right. We have a point at (1,0). (iii). basically means that the limit of this
function as x approaches 0 from the right side must be either positive or negative infinity. (iv). requires this
function to be negative for 0<x<1 which is indeed the case here.
25.
(i) f (3)  f (0)  f (2)  0
(ii) limx 2 f  x     and limx 2 f  x    
(iii) limx 1 f  x   
Step 1
1 of 3
Before drawing a graph there are three points to consider:
1. The graph crosses the x-axis in x= -3 , 0 and 2 .
2)There are 2 vertical asymptotes , x=−2and x=1
3)the graph of function increases without bound as x approaches 1 from both the left and right. Also, it
decreases without bound as x approached -2 from right and increases without bound as x approaches -2
from left .
26.
(i) f (1)  0, f (0)  1, f (1)  0
(ii) limx 1 f  x   0 and limx 1 f  x   
(iii) limx 1 f  x   1and limx 1 f  x   
This is one possible solution, answers may vary.
(i). The function must be defined at the points (-1,0), (0,1), (1,0); holes might need to be used to do this.
(ii). The function as x approaches -1 from the LEFT must approach 0, and must increase infinitely as x
approaches -1 from the right.
(iii). The function as x approaches 1 from the LEFT must approach 1, and must increase infinitely as x
approaches 1 from the RIGHT.