12. limits — ગણિત ધોરણ 11 સાયન્સ — Question

$ = \mathop {\lim }\limits_{x \to 0 + } \,\frac{{n\,{{(\log x)}^{n - 1} \frac{1}{x}}}}{{ - m{x^{ - m-1}}}}\,$    (By $L-$ Hospital's rule)

$ = \mathop {\lim }\limits_{x \to 0 + } \,\frac{{n\,{{(\log x)}^{n - 1}}}}{{ - m{x^{ - m}}}}\,$        $\left( {{\rm{Form}} \,\, \frac{\infty }{\infty }} \right)$

$ = \mathop {\lim }\limits_{x \to 0 + } \,\frac{{n\,(n - 1)\,{{(\log x)}^{(n - 2)}}\frac{1}{x}}}{{{{( - m)}^2}{x^{ - m - 1}}}}$        (By $L-$ Hospital's rule)

$ = \mathop {\lim }\limits_{x \to 0 + } \,\frac{{n\,(n - 1)\,{{(\log x)}^{n - 2}}}}{{{m^2}{x^{ - m}}}}\,$       $\left( {{\rm{Form}} \,\, \frac{\infty }{\infty }} \right)$
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$ = \mathop {\lim }\limits_{x \to 0 + } \,\frac{{n\,\,!}}{{{{( - m)}^n}{x^{ - m}}}} = 0$

(Differentiating ${N^r}$ and ${D^r}$   $n$ times).