Q. 95 - Calculus

Question

Use logarithms to prove that for any real number r,

$\underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}{\left(1+\frac{r}{x}\right)}^{\mathbf{x}}\mathbf{=}{\mathit{e}}^{\mathbf{r}}$

Step-by-Step Solution

Answer

Ans:  For real number r we con use the logarithmic eq as $\ln y=\underset{x\to \infty }{\lim }\frac{\frac{1}{1+\frac{r}{x}}·\left(-\frac{r}{x^2}\right)}{-\frac{1}{x^2}}$ 

Step 1. Given Information:

$\underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}{\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{r}}{\mathbf{x}}\mathbf{)}}^{\mathbf{x}}\mathbf{=}{\mathit{e}}^{\mathbf{r}}$

Step 2. Evaluating the values to prove:

$$\begin{gathered} let, y=\underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}{\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{r}}{\mathbf{x}}\mathbf{)}}^{\mathbf{x}}\mathbf{=}{\mathit{e}}^{\mathbf{r}}\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ Apply natural \log arithm on both sides of the above eq \\ \ln y=\ln \left( \underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}{\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{r}}{\mathbf{x}}\mathbf{)}}^{\mathbf{x}}\right) \mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \\ Since the limit and \log arithm function are interchangeable \\ \ln y=\ln \left( \underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}{\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{r}}{\mathbf{x}}\mathbf{)}}^{\mathbf{x}}\right) \\ = \underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\left(x\left(\ln \mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{r}}{\mathbf{x}}\mathbf{)}\right)\right) (\sin ce \ln {\left(x\right)}^n=n\ln (x\left)\right) \\ =\underset{x\to \infty }{\lim }\frac{\ln \mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{r}}{\mathbf{x}}\mathbf{)}}{{\displaystyle \frac{1}{x}}} \\ The right hand side of the above eq is in \frac{0}{0} form x\to \infty \end{gathered}$$

Step 3. Applying L' Hospital rule

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to \infty }{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)} (provided \frac{f(\infty )}{g(\infty )}=\frac{0}{0}and g\text{'}(\infty )=0) \\ \ln y=\underset{x\to \infty }{\lim }\frac{\frac{1}{1+\frac{r}{x}}·\left(-\frac{r}{x^2}\right)}{-\frac{1}{x^2}} \\ = \underset{x\to \infty }{\lim }\frac{\frac{x}{x+r}·\left(\frac{r}{x^2}\right)}{\frac{1}{x^2}} \\ = \underset{x\to \infty }{\lim }\frac{x}{x+r}·x^2·\left(\frac{r}{x^2}\right) \\ = \underset{x\to \infty }{\lim }\frac{xr}{x+r} \end{gathered}$$

Step 4. Rewriting the logarithmic into the exponential form:

Dividing both the numerator and the denominator by x on the right-hand side of the above equation.

$$\begin{gathered} \underset{ \mathrm{x}\to \infty }{\ln y = \lim } \frac{r}{1+\frac{\mathrm{r}}{\mathrm{x}}} \\ =\frac{r}{1+\left(\frac{\mathrm{r}}{\infty }\right)} (\sin ce \frac{1}{\infty }=0) \\ =0 \\ \mathrm{Rewriting} \mathrm{the} \mathrm{above} \mathrm{logarithmic} \mathrm{into} \mathrm{the} \mathrm{exponential} \mathrm{form}, \\ y={\mathrm{e}}^r \\ Thus,\underset{x\to \infty }{\lim } {\left(1+\frac{\mathrm{r}}{\mathrm{x}}\right)}^x=\mathrm{er} (\mathrm{from} \mathrm{eq} 1) \\ \therefore \mathrm{for} \mathrm{any} \mathrm{real} \mathrm{no} \mathrm{r} ,\underset{\mathrm{x}\to \infty }{\lim } {\left(1+\frac{\mathrm{r}}{\mathrm{x}}\right)}^{\mathrm{x}}={\mathrm{e}}^r \end{gathered}$$