6.28
Find any vertical and horizontal asymptotes of the graph of the function f(x) = (4x - 5) /(3x + 2).
Remember that a vertical asymptote is a vertical line x = c to which the graph gets closer and closer as x
approaches c from the right or from the left. Hence, we obtain vertical asymptotes by setting the denominator
3x + 2 = 0. Thus, the only vertical asymptote is the line x=-\. Recall that a horizontal asymptote is a line
y = d to which the graph gets closer and closer as x—»+«> or x—»—«. In this case,
Thus, the line y = § is a horizontal asymptote both on the right and the left.
6.29
Find the vertical and horizontal asymptotes of the graph of the function f(x) = (2x + 3) Nx
2 - 2x - 3.
x
2 — 2x — 3 = (x — 3)(x + 1). Hence, the denominator is 0 when x = 3 and when x = — 1. So, those
lines are the vertical asymptotes. (Observe that the numerator is not 0 when x — 3 and when x — — 1.) To
obtain horizontal asymptotes, we compute
divide numerator and denominator by A:. The first limit becomes
6.30
Find the vertical and horizontal asymptotes of the graph of the function f(x) = (2x + 3) /Vx
2 - 2x + 3.
Completing the square: x
2 — 2x + 3 = (x — I)
2 + 2. Thus, the denominator is always positive, and therefore , there are no vertical asymptotes. The calculation of the horizontal asymptotes is essentially the same as that
in Problem 6.29; y = 2 is a horizontal asymptote on the right and y = — 2 a horizontal asymptote on the
left.
6.33
Find the one-sided limits lim f(x) and lim f(x) if
(See Fig. 6-2.) As x approaches 2 from the right, the value f(x) = 7x-2 approaches 7(2)-2 =
14 - 2 = 12. Thus, lim f(x) = 12. As x approaches 2 from the left, the value f(x) = 3x + 5 approaches
3(2) + 5 = 6 + 5 = 11. ""Thus, lim f(x) = 11.
39
LIMITS
For the second limit, remember that
when
Hence, the horizontal asymptotes are y = 2 on the right and y = -2 on the left.
6.31
6.32
Find the vertical and horizontal asymptotes of the graph of the function f(x) = Vx + 1 - Vx.
The function is defined only for x>0. There are no values x = c such that f(x) approaches ocas x—»c.
So, there are no vertical asymptotes. To find out whether there is a horizontal asymptote, we compute
lim VTTT - Vx:
Thus, y = 0 is a horizontal asymptote on the right.
Find the vertical and horizontal asymptotes of the graph of the function /(*) = (x
2 - 5x + 6) /(x - 3).
x
2 - 5x + 6 = (x - 2)(x - 3). So, (x
2 - 5x + 6)l(x - 3) = x - 2. Thus, the graph is a straight line =
x-2 [except for the point (3,1)], and, therefore, there are neither vertical nor horizontal asymptotes.
and
In both cases, we
Find any vertical and horizontal asymptotes of the graph of the function f(x) = (4x - 5) /(3x + 2).
Remember that a vertical asymptote is a vertical line x = c to which the graph gets closer and closer as x
approaches c from the right or from the left. Hence, we obtain vertical asymptotes by setting the denominator
3x + 2 = 0. Thus, the only vertical asymptote is the line x=-\. Recall that a horizontal asymptote is a line
y = d to which the graph gets closer and closer as x—»+«> or x—»—«. In this case,
Thus, the line y = § is a horizontal asymptote both on the right and the left.
6.29
Find the vertical and horizontal asymptotes of the graph of the function f(x) = (2x + 3) Nx
2 - 2x - 3.
x
2 — 2x — 3 = (x — 3)(x + 1). Hence, the denominator is 0 when x = 3 and when x = — 1. So, those
lines are the vertical asymptotes. (Observe that the numerator is not 0 when x — 3 and when x — — 1.) To
obtain horizontal asymptotes, we compute
divide numerator and denominator by A:. The first limit becomes
6.30
Find the vertical and horizontal asymptotes of the graph of the function f(x) = (2x + 3) /Vx
2 - 2x + 3.
Completing the square: x
2 — 2x + 3 = (x — I)
2 + 2. Thus, the denominator is always positive, and therefore , there are no vertical asymptotes. The calculation of the horizontal asymptotes is essentially the same as that
in Problem 6.29; y = 2 is a horizontal asymptote on the right and y = — 2 a horizontal asymptote on the
left.
6.33
Find the one-sided limits lim f(x) and lim f(x) if
(See Fig. 6-2.) As x approaches 2 from the right, the value f(x) = 7x-2 approaches 7(2)-2 =
14 - 2 = 12. Thus, lim f(x) = 12. As x approaches 2 from the left, the value f(x) = 3x + 5 approaches
3(2) + 5 = 6 + 5 = 11. ""Thus, lim f(x) = 11.
39
LIMITS
For the second limit, remember that
when
Hence, the horizontal asymptotes are y = 2 on the right and y = -2 on the left.
6.31
6.32
Find the vertical and horizontal asymptotes of the graph of the function f(x) = Vx + 1 - Vx.
The function is defined only for x>0. There are no values x = c such that f(x) approaches ocas x—»c.
So, there are no vertical asymptotes. To find out whether there is a horizontal asymptote, we compute
lim VTTT - Vx:
Thus, y = 0 is a horizontal asymptote on the right.
Find the vertical and horizontal asymptotes of the graph of the function /(*) = (x
2 - 5x + 6) /(x - 3).
x
2 - 5x + 6 = (x - 2)(x - 3). So, (x
2 - 5x + 6)l(x - 3) = x - 2. Thus, the graph is a straight line =
x-2 [except for the point (3,1)], and, therefore, there are neither vertical nor horizontal asymptotes.
and
In both cases, we