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

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(C) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } \frac{{{{(2x + 1)}^{40}}{{(4x - 1)}^5}}}{{{{(2x + 3)}^{45}}}}$,we factor out the hig

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(C) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } \frac{{{{(2x + 1)}^{40}}{{(4x - 1)}^5}}}{{{{(2x + 3)}^{45}}}}$,we factor out the highest power of $x$ from each term:$= \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}}}}$$= \mathop {\lim }\limits_{x 	o \infty } \frac{{{x^{40}}{{(2 + \frac{1}{x})}^{40}} \cdot {x^5}{{(4 - \frac{1}{x})}^5}}}{{{x^{45}}{{(2 + \frac{3}{x})}^{45}}}}$$= \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}}}}$$= \frac{{{{(2 + 0)}^{40}}{{(4 - 0)}^5}}}{{{{(2 + 0)}^{45}}}}$$= \frac{{{2^{40}} \cdot {4^5}}}{{{2^{45}}}}$$= \frac{{{2^{40}} \cdot {{(2^2)}^5}}}{{{2^{45}}}}$$= \frac{{{2^{40}} \cdot {2^{10}}}}{{{2^{45}}}}$$= \frac{{{2^{50}}}}{{{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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