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(B) Let $g(x) = x^{1/x}$. We analyze the behavior of $g(x)$ for $x \in (0, \infty)$.Taking the natural logarithm,$\ln(g(x)) = \frac{\ln(x)}{x}$.Let $h

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(B) Let $g(x) = x^{1/x}$. We analyze the behavior of $g(x)$ for $x \in (0, \infty)$.Taking the natural logarithm,$\ln(g(x)) = \frac{\ln(x)}{x}$.Let $h(x) = \frac{\ln(x)}{x}$. Then $h'(x) = \frac{1 - \ln(x)}{x^2}$.The function $h(x)$ has a maximum at $x = e$,where $h(e) = \frac{1}{e} \approx 0.367$.Thus,$g(x) = e^{h(x)}$ has a maximum at $x = e$,where $g(e) = e^{1/e} \approx e^{0.367} \approx 1.44$.As $x 	o 0^+$,$g(x) 	o 0$,and as $x 	o \infty$,$g(x) 	o 1$.The range of $g(x)$ is $(0, e^{1/e}]$.The greatest integer function $[g(x)]$ is discontinuous whenever $g(x)$ takes an integer value.Since the range is $(0, 1.44]$,the only integer value $g(x)$ can take is $1$.$x^{1/x} = 1$ implies $x = 1$.At $x = 1$,$f(1) = [1^{1/1}] = [1] = 1$.For $x$ slightly less than $1$,$x^{1/x} For $x$ slightly greater than $1$,$x^{1/x} > 1$ (but less than $1.44$),so $[x^{1/x}] = 1$.Since the left-hand limit and right-hand limit are different at $x = 1$,the function is discontinuous at $x = 1$.Thus,there is only $1$ point of discontinuity.

Let $[t]$ represent the greatest integer not more than $t$. Then the number of discontinuous points of $f(x) = [x^{1/x}]$ in $(0, \infty)$ is

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