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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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}}}} = $

Similar Questions

$\lim_{n \to \infty} \frac{(n + 2)! + (n + 1)!}{(n + 2)! - (n + 1)!}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{1/x}} - 1}}{{{e^{1/x}} + 1}} = $

$\lim _{x \rightarrow-\infty} \log _e(\cosh x)+x=$

$\mathop {\lim }\limits_{x \to {0^ + }} {\left( {\sec \sqrt x } \right)^{\frac{{10}}{x}}}$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module