8.9. EXERCISES 175

In general, the amount you would have at the end of n months is 100(1+ r)n .When a bank says they offer 6% compounded monthly, this means r, the rate per pay-

ment period equals .06/12. Consider the problem of a rate of r per year and compoundingthe interest n times a year and letting n increase without bound. This is what is meant bycompounding continuously. The interest rate per payment period is then r/n and the num-ber of payment periods after time t years is approximately tn. From the above, the amountin the account after t years is

P(

1+rn

)nt(8.31)

Recall from Example 8.8.10 that limy→∞

(1+ x

y

)y= ex. The expression in 8.31 can be

written asP[(

1+rn

)n]t

and so, taking the limit as n→ ∞, you get Pert = A. This shows how to compound interestcontinuously.

Example 8.8.11 Suppose you have $100 and you put it in a savings account which pays6% compounded continuously. How much will you have at the end of 4 years?

From the above discussion, this would be 100e(.06)4 = 127.12. Thus, in 4 years, youwould gain interest of about $27.

8.9 Exercises1. Find the limits.

(a) limx→03x−4sin3x

tan3x

(b) limx→ π2− (tanx)x−(π/2)

(c) limx→1arctan(4x−4)arcsin(4x−4)

(d) limx→0arctan3x−3x

x3

(e) limx→0+9secx−1−13secx−1−1

(f) limx→03x+sin4x

tan2x

(g) limx→π/2ln(sinx)x−(π/2)

(h) limx→0cosh2x−1

x2

(i) limx→0−arctanx+x

x3

(j) limx→0x8 sin 1

xsin3x

(k) limx→∞ (1+5x)2x

(l) limx→0−2x+3sinx

x

(m) limx→1ln(cos(x−1))

(x−1)2

(n) limx→0+ sin1x x

(o) limx→0 (csc5x− cot5x)

(p) limx→0+3sinx−12sinx−1

(q) limx→0+ (4x)x2

(r) limx→∞x10

(1.01)x

(s) limx→0 (cos4x)(1/x2)

2. Find the following limits.

(a) limx→0+1−√

cos2xsin4(4

√x).

(b) limx→02x2−25x

sin(

x25

)−sin(3x)

.

(c) limn→∞ n( n√

7−1).

(d) limx→∞

( 3x+25x−9

)x2.