The function is defined as f(x) = (1 + frac{1 | vedclass

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(A) Let $y = (1 + \frac{1}{x})^x$. Taking the natural logarithm on both sides,we get $\ln(y) = x \ln(1 + \frac{1}{x})$. As $x 	o \infty$,$\ln(y) = x(

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$(1, e)$

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(A) Let $y = (1 + \frac{1}{x})^x$. Taking the natural logarithm on both sides,we get $\ln(y) = x \ln(1 + \frac{1}{x})$. As $x 	o \infty$,$\ln(y) = x(\frac{1}{x} - \frac{1}{2x^2} + \dots) = 1 - \frac{1}{2x} + \dots 	o 1$,so $y 	o e$. As $x 	o -\infty$,$\ln(y) = x(\frac{1}{x} - \frac{1}{2x^2} + \dots) = 1 - \frac{1}{2x} + \dots 	o 1$,so $y 	o e$. As $x 	o 0^+$,$\ln(y) = x \ln(1 + \frac{1}{x}) 	o 0 \cdot \infty$. Using $L$'Hopital's rule,$\lim_{x 	o 0^+} \frac{\ln(1 + 1/x)}{1/x} = \lim_{t 	o \infty} \frac{\ln(1+t)}{t} = 0$,so $y 	o e^0 = 1$. As $x 	o -1^+$,$f(x) 	o 0^0$ (undefined/approaching $0$). By analyzing the derivative $f'(x) = f(x) [\ln(1 + \frac{1}{x}) - \frac{1}{x+1}]$,it can be shown that the function is strictly increasing on $(-\infty, -1)$ and $(0, \infty)$. The range of the function is $(1, e)$.

The function is defined as $f(x) = (1 + \frac{1}{x})^x$. What is the range of the function $f(x)$?

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