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

$(a)\;0$$\qquad(b)\;\large\frac{1}{10}$$\qquad(c)\;\large\frac{1}{5}$$\qquad(d)\;\large\frac{3}{10}$

1 Answer

$\lim\limits_{x\to \infty}\large\frac{(2x-3)(3x-4)}{(4x-5)(5x-6)}=$$\lim\limits_{x\to \infty}\large\frac{(2-3/x)(3-4/x)}{(4-5/x)(5-6/x)}$
$\Rightarrow \large\frac{(2-0)(3-0)}{(4-0)(5-0)}$
$\Rightarrow \large\frac{6}{20}$
$\Rightarrow \large\frac{3}{10}$
Hence (d) is the correct answer.
answered Dec 24, 2013 by sreemathi.v
 
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