Math 103. Homework 4. Solutions

1. Find the limits, if they exist.

(a)




lim x® ¥  x3+2x+1 1-4x3


(b)




lim x® -¥  x-1 2x3+x+2

Solution.(a) Divide both numerator and denominator by the highest power of x in the denominator, which is x³. It obtains

lim x® ¥  x3+2x+1 1-4x3 = lim x® ¥  1 x3 (x3+2x+1) 1 x3 (1-4x3) = lim x® ¥  1+ 2 x + 1 x3 -4+ 1 x3 = -1 4

(b) The same technique as that used in (a) gives

lim x® -¥  x-1 2x3+x+2 = lim x® -¥  1 x3 (x-1) 1 x3 (2x3 +x+2) = lim x® -¥  1 x2 - 1 x3 2+ 1 x2 + 2 x3 = 0-0 2+0+0 = 0 2 =0

3.[84, p. 133] The concentration of a certain drug in a patient's bloodstream t hours after injection is given by

C(t) = (0.2) t t2+1

mg/cm³. Evaluate lim_{t® ¥} C(t) and interpret your result.

Solution.

lim t®¥  C(t) = lim t®¥  (0.2)t t2+1 = lim t®¥  (0.2)t t2 1+ 1 t2 = 0 1+0 =0.

This says that the the concentration of the drug in the bloodstream disappears as time goes by.

5.[42, p. 146] Find the one-sided limits lim_{x® 1+} f(x) and lim_{x® 1-}f(x), if they exist, where f(x) is given by

f(x) = ì ï ï í ï ï î x+2   ___ Öx-1   if x ³ 1, 1-   ___ Ö1-x   if x < 1

Solution.

lim x® 1+  f(x)= lim x® 1+  x+2   ___ Öx-1   = 1+2   ___ Ö1-1   =1

In words, to find the right limit of f(x) at x=1, simply plug x=1 into the expression of f that is valid to the right of 1 (i.e., where x ³ 1).

Similarly,

lim x® 1-  f(x)= lim x® 1-  1+   ___ Ö1-x   = 1+   ___ Ö1-1   =1

4.[62, p. 147] Find the values of x for which the following function is continuous

f(x)= ì í î x+1 if x £ 1, 1-x2 if x > 1

Solution. f(x) is continuous on (-¥,1)È(1,¥). The details are as follows.

The only point where f(x) may fail to be continuous is at x=1. Away from x=1 the function f(x) is like a polynomial, and so it is continuous. To see all the work, take a ¹ 1. Then, if a > 1 you have

lim x® a  f(x) = lim x® a  1-x2 = 1-a2 = f(a)

so it is continuous at a by the very definition of continuous function. Analogously, if a < 1,

lim x® a  f(x) = lim x® a  x+1 = a+1 = f(a)

and so f is also continuous at a.

To check continuity at x=1, we first compute right and left limits:

lim x® 1+  f(x) = lim x® 1+  1-x2 = 1-(1)2 = 0

and

lim x® 1-  f(x) = lim x® 1-  x+1 = 1+1=2.

Since right and left limits are distinct, the limit lim_{x® 1} f(x) does not exist, and so f is not continuous at x=1.

5.[92, p. 154] Joan is looking straight out of a window of an apartment building at a height of 32 ft from the ground. A boy throws a tennis ball straight up by the side of the building where the window is located. The height of the ball (measures in feet) from the ground at time t is h(t) = 4+64 t - 16 t².

(a) Show that h(0)=4 and h(2)=68 and then use the intermediate value theorem to conclude that the ball must cross Joan's line of sight at least once.
(b) At what time(s) does the ball cross Joan's line of sight? (Hint. You have to solve a quadratic equation.)

Solution.(a) The values h(0)=4+64 (0) -16(0)²=4 and h(2)=4+64(2)-16(2)²=4+128-64=68. The function h(t) is continuous on the interval [0,2] because it is a polynomial. The intermediate value theorem says that, on this interval, h(t) takes on all values between h(0)=4 and h(2)=68. Joan's line of sight is at height 32, which is between 4 and 68.

(b) The ball crosses Joan's line of sight at time t if its height at this time is h(t)=32. Substituting for h gives the quadratic equation

4+64 t-16 t2=32

which is also

16 t2 -64 t +28=0

The quadratic formula gives the solutions t=1/2 and t=7/2. The ball actually crosses Joan's line of sight twice: on its way up and on its way down.