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Which one of the following is not bounded on the intervals as indicated
$f(x) =$ ${2^{\frac{1}{{x - 1}}}}$ on $(0, 1)$
$g(x) = x cos \frac{1}{x} $ on $(-\infty ,\infty )$
$h(x) = xe^{-x} $ on $(0, \infty )$
$l (x) =tan^{-1} 2^x $ on $ (-\infty , \infty )$
Solution
$(A)$ $\mathop {Limit}\limits_{x \to {0^ + }} \,$ $f (x) = \mathop {Limit}\limits_{h \to 0} \,{2^{\frac{1}{{h – 1}}}}\, = \,\frac{1}{2}\,$ ;
$\mathop {Limit}\limits_{x \to {1^ – }} \,$ $f (x) = \,\mathop {Limit}\limits_{h \to 0} \,{2^{\frac{1}{{ – h}}}}\, = \,0$
$\Rightarrow$ $\,\,f\,(x)\, \in \,\left( {0,\,\frac{1}{2}} \right)\,$ $\Rightarrow$ bounded
$(C)$ $\mathop {Limit}\limits_{h \to 0} \,$ $x e^{-x} =$ $\mathop {Limit}\limits_{h \to 0} \,$ $h e^{ -h} = 0$ ;
$\mathop {Limit}\limits_{x \to \infty } \,$ $x e^{-x} = $$\mathop {Limit}\limits_{x \to \infty } \,$ $\frac{x}{{{e^x}}}\, = \,0$
$\Rightarrow$ Also $y\, = \,\frac{x}{{{e^x}}}$ $\Rightarrow$ $y’ =$$\frac{{{e^x}\, – \,x{e^x}}}{{{e^{2x}}}}\,$ $e ^{x (1 – x)}$ $\Rightarrow$ $h(x)\, = \,\left( {0\,,\,\frac{1}{e}} \right]$
