Question

If an initial amount A0 of money is invested at an interest rate i compounded times a year, the value of the investment after t years is If we let n = 8, that is referred to as the continuous compounding of interest. Use L' Hospital's Rule to show that if interest is compounded continuously, then the amount after years is A=A0eit

Answer 1

Answer:

Following are the solution to the given point:

Step-by-step explanation:

Please find the comp[lete question in the attached file.

Given:

$\bold{ \lim_{n \to \ \infty} (1+ \frac{r}{n})^{nt} =e^{rt}}$

In point 1:

$\to y = (1+ \frac{r}{n})^{nt}$

In point 2:

$\to \ln (y)= nt \ln(1+ \frac{r}{n})$

In point 3:

Its key thing to understand, which would be that you consider the limit n to$\infty$,  in which r and t were constants!  

$=lim_{n \to \ \infty} \ln (y) = lim_{n \to \ \infty} nt \ln(1+\frac{r}{n})\\\\= lim_{n \to \ \infty} \frac{\ln(1+\frac{r}{n})}{\frac{1}{nt}}\\\\= lim_{n \to \ \infty} \frac{\frac{-r}{\frac{n^2}{(1+\frac{r}{n})}}}{- \frac{1}{n^2t}}\\\\= lim_{n \to \ \infty} \frac{\frac{rn^2t}{n^2}}{(1+\frac{r}{n})}\\\\= lim_{n \to \ \infty} \frac{rt}{(1+\frac{r}{n})}\\\\= \frac{rt}{(1+\frac{r}{0})}\\\\=rt$

In point 4:

$\to \lim_{n \to \ \infty} = (1+\frac{r}{n})^{nt}$ and

$\to \lim_{n \to \ \infty} = A_0e^{rt}$