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

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(D) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } \frac{(2x - 3)(3x - 4)}{(4x - 5)(5x - 6)}$,we divide the numerator and the denominat

Vedclass · Accepted Answer

$\frac{3}{10}$

Vedclass · Answer

(D) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } \frac{(2x - 3)(3x - 4)}{(4x - 5)(5x - 6)}$,we divide the numerator and the denominator by $x^2$:$= \mathop {\lim }\limits_{x 	o \infty } \frac{x(2 - \frac{3}{x}) \cdot x(3 - \frac{4}{x})}{x(4 - \frac{5}{x}) \cdot x(5 - \frac{6}{x})}$$= \mathop {\lim }\limits_{x 	o \infty } \frac{(2 - \frac{3}{x})(3 - \frac{4}{x})}{(4 - \frac{5}{x})(5 - \frac{6}{x})}$As $x 	o \infty$,the terms $\frac{3}{x}, \frac{4}{x}, \frac{5}{x}, \frac{6}{x}$ approach $0$:$= \frac{(2 - 0)(3 - 0)}{(4 - 0)(5 - 0)} = \frac{2 	imes 3}{4 	imes 5} = \frac{6}{20} = \frac{3}{10}$

$\mathop {\lim }\limits_{x \to \infty } \frac{{(2x - 3)(3x - 4)}}{{(4x - 5)(5x - 6)}} = $

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