Evaluate the following limits, where c and k are constants.

$\underset{x\to 2}{lim}(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x})$

Ava-May Nelson
2021-10-17
Answered

Evaluate the following limits, where c and k are constants.

$\underset{x\to 2}{lim}(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x})$

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wheezym

Answered 2021-10-18
Author has **103** answers

To evaluate: $\underset{x\to 2}{lim}(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x})$

Solution:

Given,$\underset{x\to 2}{lim}(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x})$

On simplifying further we get:

$\underset{x\to 2}{lim}(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x})=\underset{x\to 2}{lim}(\frac{1}{x-2}-\frac{2}{x(x-2)})$

$\Rightarrow \underset{x\to 2}{lim}(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x})=\underset{x\to 2}{lim}\left(\frac{x-2}{x(x-2)}\right)$

$\Rightarrow \underset{x\to 2}{lim}(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x})=\underset{x\to 2}{lim}\frac{1}{x}$

$\Rightarrow \underset{x\to 2}{lim}(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x})=\frac{1}{2}$

Result:

$\underset{x\to 2}{lim}(\frac{1}{x-2}-\frac{2}{{x}^{2}-2x})=\frac{1}{2}$

Solution:

Given,

On simplifying further we get:

Result:

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