A graphing calculator is recommended.If you deposit $1 into a bank account paying 12% interest compounded continuously, a year later its value will be  lim x→0 (1+3x/25)^1/x Find the limit by making a TABLE of values correct to two decimal places, thereby finding the value of the deposit in dollars and cents. Values in table from -0.1 to 0.0001    

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Since the interest is compounded continuously, we know the answer is given by Pe^(rt) where P is the pricipal (here 1), e is Euler's constant (e approximately 2.71828), r is the annual interest rate, and t the amount of time in years.

So the answer will be e^(.12) which is approximately 1.127496852 or 1.13.

To find the limit, we need L'Hopital's rule. Please see the attachment:

 

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`lim_(x->0) (1+(3x)/25))^(1/x)`

To solve, let's assign values to x that are approaching zero from the left and from the right.

For x values that are approaching zero from the left:

`x=-0.1`

`y= (1+(3(-0.1))/25)^(1/(-0.1))=1.128315499=1.13`

`x= -0.05`

`y= (1+(3(-0.1))/25)^(1/(-0.01))=1.127904455=1.13`

`x=-0.002`

`y=(1+(3(-0.002))/25)^(1/(-0.002))=1.127513090=1.13`

For x values that are approaching zero from the right:

`x=0.0001`

`y=(1+(3(0.0001))/25)^(1/0.0001)=1.127496039=1.13`

`x=0.00005`

`y=(1+(3(0.00005))/25)^(1/0.00005)=1.127496445=1.13`

`x=0.000009`

`y=(1+(3(0.000009))/25)^(1/0.000009))=1.127496778=1.13`

(See attachment for the table.)

 

Therefore, `lim_(x->0) (1+(3x)/25)^(1/x) = 1.13`   .

 

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