લક્ષની કિંમત શોધો: lim _{x rightarrow infty}l | vedclass

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Question

(B) ધારો કે $L = \lim _{x}$ ${\rightarrow \infty}\left\{x-\left(\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)\right)^{1/n}\right\}$.$n

Vedclass · Accepted Answer

$\frac{a_1+a_2+\ldots+a_n}{n}$ છે

Vedclass · Answer

(B) ધારો કે $L = \lim _{x}$ ${\rightarrow \infty}\left\{x-\left(\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)\right)^{1/n}\right\}$.$n$-th મૂળની અંદરના પદમાંથી $x$ સામાન્ય લેતા:$L = \lim _{x}$ ${\rightarrow \infty} \left\{ x - x \left( \left(1-\frac{a_1}{x}\right)\left(1-\frac{a_2}{x}\right) \ldots\left(1-\frac{a_n}{x}\right) \right)^{1/n} \right\}$.દ્વિપદી અંદાજ $(1-u)^k \approx 1-ku$ નો ઉપયોગ કરતા:$L = \lim _{x}$ ${\rightarrow \infty} x \left\{ 1 - \left(1-\frac{a_1}{nx}\right)\left(1-\frac{a_2}{nx}\right) \ldots\left(1-\frac{a_n}{nx}\right) \right\}$.ગુણાકારનું વિસ્તરણ કરતા અને $O(1/x)$ સુધીના પદો રાખતા:$L = \lim _{x}$ ${\rightarrow \infty} x \left\{ 1 - \left( 1 - \frac{a_1+a_2+\ldots+a_n}{nx} + O\left(\frac{1}{x^2}\right) \right) \right\}$.$L = \lim _{x \rightarrow \infty} x \left( \frac{a_1+a_2+\ldots+a_n}{nx} \right) = \frac{a_1+a_2+\ldots+a_n}{n}$.

લક્ષની કિંમત શોધો: $\lim _{x \rightarrow \infty}\left\{x-\sqrt[n]{\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)}\right\}$,જ્યાં $a_1, a_2, \ldots, a_n$ ધન સંમેય સંખ્યાઓ છે.

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