Range and Graph of a function

Question

Let $$f(x)=\left(1+\frac{1}{x}\right)^x$$

State the range and plot its graph wherever the function is defined.

I got the range and $f'(x)$, but I cannot find the sign of it to predict the increasing/decreasing of the graph. Please help.

Answer 1

The domain of this function $f(x)=(1+1/x)^x$ is $(-\infty, -1) \cup (0,\infty)$ Range of this function is $(1,\infty)$. For this $$L_1=\lim_{x \to 0} (1+1/x)^x= \exp[\lim_{x \to 0} x \ln (1+1/x)]$$ By L-Hospital $$L_1=\lim_{x \to 0}\exp[ \frac{\ln(1+1/x)}{1/x}]=\exp[ \lim_{x \to 0} \frac{x}{1+x}]=1.$$

Next $$L_2=\lim_{x \to \pm\infty} (1+1/x)^x= \exp[\lim_{x \to \pm\infty} x \ln (1+1/x)]$$ $$L_2=\lim_{x \to \pm\infty}\exp[ \frac{\ln(1+1/x)}{1/x}]=$$ By L-Hospital $$L_2=\lim_{x \to \pm\infty}\exp[ \frac{\ln(1+1/x)}{1/x}]=\exp[ \lim_{x \to \pm\infty} \frac{x}{1+x}]=e.$$

Next $$L_3=\lim_{x \to -1^{-}} \ln (1+1/x)^x=\lim_{h \to 0} [1+1/(-1-h)^{-1-h}=\lim_{h\to 0} (1+1/h)^{1+h}=\infty$$

Additionally, $f(x)=(1+1/x)^x$ is an increasing function in its domain. See $$f'(x)=(1+1/x)e^{-1/(1+x)}>0, x \in (-\infty, -1) \cup (0, \infty)$$, see the graph of $f(x)$ below.

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Answer 2

Hint

$$f(x)=\left(1+\frac{1}{x}\right)^x\implies \log(f(x))=x \log\left(1+\frac{1}{x}\right)$$ $$\frac{f'(x)}{f(x)}=\log \left(1+\frac{1}{x}\right)-\frac{1}{x+1}$$ Now, play with some inequalities.