Question

1. eric deposits $4,700 at 1.03% interest, compounded continuously for five years.

a. what is his ending balance?
b. how much interest did the account earn?

1. write the verbal sentence that is the translation of $lim_{x \to +infty} f(x)=3.66$.
2. write the verbal sentence given below symbolically using limit notation.

the limit of g(x), as x approaches zero, is fifteen.

1. given the function $f(x)=\frac{2x - 17}{x}$, use a table to find $lim_{x \to +infty} f(x)$.
2. find the balance for each compounding period on $50,000 for $2\frac{1}{2}$ years at a rate of 1.3%.

a. annually = 51640.88
b. semiannually = 51,646.26
c. quarterly = 51,642.97
d. monthly = 51,650.74
e. daily
f. hourly = 51,651.69
g. continuously

1. a private university has an endowment fund that currently has 49 million dollars in it. if it is invested in a one - year cd that pays 2% interest compounded continuously, how much interest will it earn?
2. use a table of increasing values of x to find each of the following limits.

a. $lim_{x \to +infty} f(x)$ if $f(x)=\frac{5x - 2}{x + 3}=5$
b. $lim_{x \to +infty} g(x)$ if $g(x)=\frac{12x + 5}{4x+3}=3$
c. $lim_{x \to +infty} f(x)$ if $f(x)=\frac{5x^{3}-100}{x^{2}}=+infty$
d. $lim_{x \to +infty} f(x)$ if $f(x)=\frac{7x^{2}-1}{x^{3}+2}=0$

1. find the interest earned on a $14,000 balance for nine months at 1.1% interest compounded continuously.
2. assume you had p dollars to invest in an account that paid 5% interest compounded continuously. how long would it take your money to double? (hint: try substituting different numbers of years into the continuous compounding formula). round to the nearest year.

34 financial algebra workbook 2 - 6

Explanation:

13.

a.

Step1: Recall continuous - compounding formula

The formula for continuous compounding is $A = Pe^{rt}$, where $P$ is the principal amount, $r$ is the annual interest rate (in decimal form), $t$ is the time in years, and $A$ is the ending balance.
Given $P=\$4700$, $r = 0.0103$ (since $1.03\%=0.0103$), and $t = 5$.

Step2: Substitute values into the formula

$A=4700\times e^{0.0103\times5}=4700\times e^{0.0515}$.
Using a calculator, $e^{0.0515}\approx1.0528$.
So $A = 4700\times1.0528=\$4958.16$.

Step1: Recall the interest - earning formula

The interest $I$ earned is $I=A - P$.
We know $A = 4958.16$ and $P = 4700$ from part (a).

Step2: Calculate the interest

$I=4958.16−4700=\$258.16$.

The limit notation $\lim_{x
ightarrow+\infty}f(x)=3.66$ can be translated as "The limit of the function $f(x)$ as $x$ approaches positive infinity is $3.66$".

Answer:

$\$4958.16$

b.