અહી $ f(x) = x^{\alpha} logx$  એ $x \in (0, 1) $ માં વિકલનીય છે.

$f(x) = x^{\alpha} logx$  એ $x \in [0, 1]$  માં સતત છે.

$= x^{\alpha } logx $ એ $x \in [0, 1]$  માં જમણી બાજુ સતત છે. જેથી $ f(0) = 0 $ થાય.

હવે, ${f}{\rm{(0}}\, + \,{\rm{h)}}\, = \,\begin{array}{*{20}{c}}
{\lim }\\
{h \to 0}
\end{array}{h^\alpha }\,\log h$

$\alpha \, = \,{\rm{ – }}\,{\rm{2}}\,$ માટે $\begin{array}{*{20}{c}}
{\lim }\\
{h \to 0}
\end{array}\,{h^{ – 2}}\,\log h\, = \,\begin{array}{*{20}{c}}
{\lim }\\
{h \to 0}
\end{array}\,\frac{{\log \,h}}{{{h^2}}}\,\, = \,\,\infty $

$\alpha \,\, = \,\, – \,\,1\,$ માટે $\begin{array}{*{20}{c}}
{\lim }\\
{h \to 0}
\end{array}\,\,{h^{ – 1}}\log h\,\, = \,\,\begin{array}{*{20}{c}}
{\lim }\\
{h \to 0}
\end{array}\,\frac{{\log \,h}}{h}\,\, = \,\,\infty $

$\alpha \, = \,0$ માટે $\begin{array}{*{20}{c}}
{\lim }\\
{h \to 0}
\end{array}\,{h^0}\,\log h\, = \,1.\infty \,\, = \,\,\infty $

$\alpha \, = \,\,\frac{1}{2}$ માટે $\begin{array}{*{20}{c}}
{\lim }\\
{h \to 0}
\end{array}\,{h^{1/2}}\,\log h\,(0\, \times \,\infty $ સ્વરૂપ $)$

 $ = \,\begin{array}{*{20}{c}}
{\lim }\\
{h \to 0}
\end{array}\,\frac{{\log h}}{{{h^{ – 1/2}}}}\,\, = \,\begin{array}{*{20}{c}}
{\lim }\\
{h \to 0}
\end{array}\,\,\frac{{1/h}}{{ – \frac{1}{2}{h^{ – 3/2}}}}\,\, = \,\,\begin{array}{*{20}{c}}
{\lim }\\
{h \to 0}
\end{array}\, – \,\,\frac{{2{h^{3/2}}}}{h}\,\, = \,\begin{array}{*{20}{c}}
{\lim }\\
{\,h \to 0}
\end{array}\, – 2{h^{1/2}}\, = \,\,0$