mathop {lim }limits_{n to infty } {left( {fra | vedclass

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(B) ધારો કે $L = \mathop {\lim }\limits_{n 	o \infty } {\left( {\frac{{{n^2} - n + 1}}{{{n^2} - n - 1}}} ight)^{n(n - 1)}}$.આપણે પદાવલિને $L = \mat

Vedclass · Accepted Answer

$e^2$

Vedclass · Answer

(B) ધારો કે $L = \mathop {\lim }\limits_{n 	o \infty } {\left( {\frac{{{n^2} - n + 1}}{{{n^2} - n - 1}}} ight)^{n(n - 1)}}$.આપણે પદાવલિને $L = \mathop {\lim }\limits_{n 	o \infty } {\left( {\frac{{(n^2 - n - 1) + 2}}{{n^2 - n - 1}}} ight)^{n(n - 1)}} = \mathop {\lim }\limits_{n 	o \infty } {\left( {1 + \frac{2}{{n^2 - n - 1}}} ight)^{n(n - 1)}}$ તરીકે લખી શકીએ.પ્રમાણિત લક્ષના સૂત્ર $\mathop {\lim }\limits_{x 	o \infty } {(1 + \frac{a}{x})^x} = e^a$ નો ઉપયોગ કરતા:$L = \exp \left( \mathop {\lim }\limits_{n 	o \infty } \frac{2}{n^2 - n - 1} \cdot n(n - 1) ight)$.$L = \exp \left( \mathop {\lim }\limits_{n 	o \infty } \frac{2(n^2 - n)}{n^2 - n - 1} ight)$.અંશ અને છેદને $n^2$ વડે ભાગતા,આપણને $\exp(2) = e^2$ મળે છે.

$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{{n^2} - n + 1}}{{{n^2} - n - 1}}} \right)^{n(n - 1)}} = $

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