For the function f(x) = (1 + frac{1}{x})^x,wh | vedclass

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(B) We evaluate each limit for the function $f(x) = (1 + \frac{1}{x})^x$: $1$. For $A$: $\lim_{x 	o \infty} (1 + \frac{1}{x})^x = e$. $2$. For $B$: $

Vedclass · Accepted Answer

$\lim_{x 	o 0^+} f(x)$

Vedclass · Answer

(B) We evaluate each limit for the function $f(x) = (1 + \frac{1}{x})^x$: $1$. For $A$: $\lim_{x 	o \infty} (1 + \frac{1}{x})^x = e$. $2$. For $B$: $\lim_{x 	o 0^+} (1 + \frac{1}{x})^x$. As $x 	o 0^+$,$\frac{1}{x} 	o \infty$,so the base $(1 + \frac{1}{x}) 	o \infty$. Thus,$\infty^0$ is an indeterminate form. Let $y = (1 + \frac{1}{x})^x$,then $\ln y = x \ln(1 + \frac{1}{x}) = x \ln(\frac{x+1}{x}) = x [\ln(x+1) - \ln x]$. As $x 	o 0^+$,$x \ln x 	o 0$,so $\ln y 	o 0 \cdot \ln(1) = 0$,thus $y 	o e^0 = 1$. $3$. For $C$: $\lim_{x 	o -1^-} (1 + \frac{1}{x})^x$. As $x 	o -1^-$,$(1 + \frac{1}{x}) 	o 0^+$,so $0^{-1} = \infty$. $4$. For $D$: $\lim_{x 	o -\infty} (1 + \frac{1}{x})^x = \lim_{t 	o \infty} (1 - \frac{1}{t})^{-t} = [\lim_{t 	o \infty} (1 - \frac{1}{t})^t]^{-1} = (e^{-1})^{-1} = e$. Therefore,the limit that tends to unity is $B$.

For the function $f(x) = (1 + \frac{1}{x})^x$,which of the following limits tends to unity?

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