As $x\to \infty$, ${\left(\frac{x}{1\+x}\right)}^x$ tends to the indeterminate form $1^{\infty }$.

From the graph of ${\left(\frac{x}{1\+x}\right)}^x$ in Figure 3.9.11(a), it would appear that the required limit is approximately 0.4. Indeed,

$\underset{x\to \infty }{lim}{\left(\frac{x}{1\+x}\right)}^x$ = ${\ⅇ}^{-1}$$\doteq 0.368$

Annotated stepwise solution via the Context Panel

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$\underset{x\to \infty }{lim}{\left(\frac{x}{1\+x}\right)}^x$$\stackrel{\text{show solution steps}}{\to }$$\begin{array}{lll}& & \text{Limit Steps}\\ & & \underset{x\to \mathrm{\infty }}{lim}{\left(\frac{x}{1+x}\right)}^x\\ \text{▫}& & \text{1. Rewrite}\\ & \text{◦}& \text{Equivalent expression}\\ & & {\left(\frac{x}{1+x}\right)}^x={\ⅇ}^{x\ln \left(\frac{x}{1+x}\right)}\\ & & \text{This gives:}\\ & & \underset{x\to \mathrm{\infty }}{lim}{\ⅇ}^{x\ln \left(\frac{x}{1+x}\right)}\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{exponential}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{exponential}}\text{rule}\\ & & \underset{x\to a}{lim}{e}^{f\left(x\right)}=e^{\underset{x\to a}{lim}f\left(x\right)}\\ & & \text{This gives:}\\ & & {\ⅇ}^{\underset{x\to \mathrm{\infty }}{lim}x\ln \left(\frac{x}{1+x}\right)}\\ \text{▫}& & \text{3. Apply the}\mathbf{\text{L'Hôpital's Rule}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{L'Hôpital's Rule}}\text{rule}\\ & & \underset{x\to c}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to c}{lim}\frac{\frac{\ⅆ}{\ⅆx}f\left(x\right)}{\frac{\ⅆ}{\ⅆx}g\left(x\right)}\\ & \text{◦}& \text{Rule applied}\\ & & x\ln \left(\frac{x}{1+x}\right)=-\frac{x}{1+x}\\ & & \text{This gives:}\\ & & {\ⅇ}^{\underset{x\to \mathrm{\infty }}{lim}\left(-\frac{x}{1+x}\right)}\\ \text{▫}& & \text{4. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\underset{x\to \mathrm{\infty }}{lim}\left(-\frac{x}{1+x}\right)\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \underset{x\to \mathrm{\infty }}{lim}Cf\left(x\right)=C\left(\underset{x\to \mathrm{\infty }}{lim}f\left(x\right)\right)\\ & \text{◦}& \text{This means:}\\ & & \underset{x\to \mathrm{\infty }}{lim}\left(-\frac{x}{1+x}\right)=\left(\mathrm{-1}\right)\cdot \left(\underset{x\to \mathrm{\infty }}{lim}\frac{x}{1+x}\right)\\ & & \text{We can rewrite the limit as:}\\ & & {\ⅇ}^{-\left(\underset{x\to \mathrm{\infty }}{lim}\frac{x}{1+x}\right)}\\ \text{▫}& & \text{5. Apply the}\mathbf{\text{L'Hôpital's Rule}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{L'Hôpital's Rule}}\text{rule}\\ & & \underset{x\to c}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to c}{lim}\frac{\frac{\ⅆ}{\ⅆx}f\left(x\right)}{\frac{\ⅆ}{\ⅆx}g\left(x\right)}\\ & \text{◦}& \text{Rule applied}\\ & & \frac{x}{1+x}=1\\ & & \text{This gives:}\\ & & {\ⅇ}^{-\left(\underset{x\to \mathrm{\infty }}{lim}1\right)}\\ \text{▫}& & \text{6. Apply the}\mathbf{\text{constant}}\text{rule to the term}\underset{x\to \mathrm{\infty }}{lim}1\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant}}\text{rule}\\ & & \mathrm{Limit}\left(C\,x\right)=C\\ & \text{◦}& \text{This means}\\ & & \underset{x\to \mathrm{\infty }}{lim}1=1\\ & & \text{We can now rewrite the limit as:}\\ & & {\ⅇ}^{\mathrm{-1}}\end{array}$