\[\lim_{x \to 2} \left( \frac{x}{x - 2} - \frac{4}{x^2 - 2x} \right)\]

\[\lim_{x \to 2} \left[ \frac{x}{x - 2} - \frac{4}{x^2 - 2x} \right]\]\[ = \lim_{x \to 2} \left[ \frac{x}{x - 2} - \frac{4}{x\left( x - 2 \right)} \right]\]\[ = \lim_{x \to 2} \left[ \frac{x^2 - 4}{x\left( x - 2 \right)} \right]\]\[ = \lim_{x \to 2} \left[ \frac{\left( x - 2 \right)\left( x + 2 \right)}{x\left( x - 2 \right)} \right]\]\[ = \lim_{x \to 2} \left[ \frac{\left( x + 2 \right)}{x} \right]\]\[ = \frac{2 + 2}{2}\]\[ = 2\]

Lim X → 2 ( X X − 2 − 4 X 2 − 2 X ) - Mathematics

Question

$lim_{x \to 2} (\frac{x}{x - 2} - \frac{4}{x^{2} - 2 x})$

Solution

$lim_{x \to 2} [\frac{x}{x - 2} - \frac{4}{x^{2} - 2 x}]$
$= lim_{x \to 2} [\frac{x}{x - 2} - \frac{4}{x (x - 2)}]$
$= lim_{x \to 2} [\frac{x^{2} - 4}{x (x - 2)}]$
$= lim_{x \to 2} [\frac{(x - 2) (x + 2)}{x (x - 2)}]$
$= lim_{x \to 2} [\frac{(x + 2)}{x}]$
$= \frac{2 + 2}{2}$
$= 2$

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Concept of Limits - Algebra of Limits

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Q 14 Q 13 Q 15

Chapter 29: Limits - Exercise 29.3 [Page 23]

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Chapter 29 Limits
Exercise 29.3 | Q 14 | Page 23

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Lim X → 2 ( X X − 2 − 4 X 2 − 2 X ) - Mathematics

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