mathop {lim }limits_{x to infty } frac{{{{(2x | vedclass

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(C) લક્ષ $\mathop {\lim }\limits_{x 	o \infty } \frac{{{{(2x + 1)}^{40}}{{(4x - 1)}^5}}}{{{{(2x + 3)}^{45}}}}$ ની ગણતરી કરવા માટે,આપણે દરેક પદમાંથી $

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(C) લક્ષ $\mathop {\lim }\limits_{x 	o \infty } \frac{{{{(2x + 1)}^{40}}{{(4x - 1)}^5}}}{{{{(2x + 3)}^{45}}}}$ ની ગણતરી કરવા માટે,આપણે દરેક પદમાંથી $x$ ની મહત્તમ ઘાત સામાન્ય કાઢીશું:$= \mathop {\lim }\limits_{x 	o \infty } \frac{{{{[x(2 + \frac{1}{x})]}^{40}}{{[x(4 - \frac{1}{x})]}^5}}}{{{{[x(2 + \frac{3}{x})]}^{45}}}}$$= \mathop {\lim }\limits_{x 	o \infty } \frac{{{x^{45}}{{(2 + \frac{1}{x})}^{40}}{{(4 - \frac{1}{x})}^5}}}{{{x^{45}}{{(2 + \frac{3}{x})}^{45}}}}$$= \frac{{{2^{40}} \cdot {4^5}}}{{{2^{45}}}} = \frac{{{2^{40}} \cdot {2^{10}}}}{{{2^{45}}}} = {2^5} = 32$

$\mathop {\lim }\limits_{x \to \infty } \frac{{{{(2x + 1)}^{40}}{{(4x - 1)}^5}}}{{{{(2x + 3)}^{45}}}} = $

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