Question #d03ef

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Question #d03ef

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Answer 1 · 2016-12-28T04:22:44

Limit $\to$ $rarr$ $(\frac{1}{x - 2} - \frac{4}{x^{3} - 2 x^{2}})$ $(1/(x-2)- 4/(x^3-2x^2 ))$
= $(\frac{1}{x - 2} - \frac{4}{x^{2} (x - 2)})$ $(1/(x-2)-4/(x^2(x-2)) )$ $n o w;$ $now;$ $(\frac{1}{x - 2})$ $(1/(x-2))$ common

hence
$\frac{1}{x - 2}$ $1/(x-2)$ $(1 - \frac{4}{x^{2}})$ $(1-4/x^2)$

$\frac{1}{x - 2}$ $1/(x-2)$ $(\frac{x^{2}}{x^{2}} - \frac{4}{x^{2}})$ $(x^2/x^2-4/x^2)$ (I took lcm here or else u can cross multiply)

now ( $a^{2} - b^{2})$ $a^2-b^2)$ = $(a - b)$ $(a-b)$ $(a + b)$ $(a+b)$

$(\frac{1}{x - 2})$ $(1/(x-2))$ $((x - 2)$ $((x-2)$ (x+2) $/ (x^{2}))$ $/(x^2))$

$(1/cancel(x-2))$ $((cancel(x-2))$ (x+2)) $/(x^2)$

now $(x+2)/x^2$

now as we cannot simplify it further

substitute limits = $((2)+2)$ / $(2)^2$

hence answer =1

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Question #d03ef

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