Which one of the following is not bounded on | vedclass

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Question

$(A)$  $\mathop {Limit}\limits_{x 	o {0^ + }} \,$ $f (x) = \mathop {Limit}\limits_{h 	o 0} \,{2^{\frac{1}{{h - 1}}}}\, = \,\frac{1}{2}\,$ ;

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$g(x) = x cos \frac{1}{x} $ on $(-\infty ,\infty )$

Vedclass · Answer

$(A)$  $\mathop {Limit}\limits_{x 	o {0^ + }} \,$ $f (x) = \mathop {Limit}\limits_{h 	o 0} \,{2^{\frac{1}{{h - 1}}}}\, = \,\frac{1}{2}\,$ ;

$\mathop {Limit}\limits_{x 	o {1^ - }} \,$ $f (x) = \,\mathop {Limit}\limits_{h 	o 0} \,{2^{\frac{1}{{ - h}}}}\, = \,0$
$\Rightarrow$  $\,\,f\,(x)\, \in \,\left( {0,\,\frac{1}{2}} ight)\,$     $\Rightarrow$ bounded 
$(C)$  $\mathop {Limit}\limits_{h 	o 0} \,$ $x e^{-x} =$ $\mathop {Limit}\limits_{h 	o 0} \,$ $h e^{ -h} = 0$ ;

$\mathop {Limit}\limits_{x 	o \infty } \,$  $x e^{-x} = $$\mathop {Limit}\limits_{x 	o \infty } \,$ $\frac{x}{{{e^x}}}\, = \,0$
$\Rightarrow$ Also $y\, = \,\frac{x}{{{e^x}}}$   $\Rightarrow$  $y&rsquo; =$$\frac{{{e^x}\, - \,x{e^x}}}{{{e^{2x}}}}\,$ $e ^{x (1 - x)}$  $\Rightarrow$  $h(x)\, = \,\left( {0\,,\,\frac{1}{e}} ight]$

Which one of the following is not bounded on the intervals as indicated

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