# Evaluate the following limits, where c and k are constants. \lim_{x\to2}(

Evaluate the following limits, where c and k are constants.
$\underset{x\to 2}{lim}\left(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x}\right)$
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wheezym
To evaluate: $\underset{x\to 2}{lim}\left(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x}\right)$
Solution:
Given, $\underset{x\to 2}{lim}\left(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x}\right)$
On simplifying further we get:
$\underset{x\to 2}{lim}\left(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x}\right)=\underset{x\to 2}{lim}\left(\frac{1}{x-2}-\frac{2}{x\left(x-2\right)}\right)$
$⇒\underset{x\to 2}{lim}\left(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x}\right)=\underset{x\to 2}{lim}\left(\frac{x-2}{x\left(x-2\right)}\right)$
$⇒\underset{x\to 2}{lim}\left(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x}\right)=\underset{x\to 2}{lim}\frac{1}{x}$
$⇒\underset{x\to 2}{lim}\left(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x}\right)=\frac{1}{2}$
Result:
$\underset{x\to 2}{lim}\left(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x}\right)=\frac{1}{2}$