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Question Number 26289 by d.monhbayr@gmail.com last updated on 23/Dec/17

$$\underset{x\to 2}{\lim }(\frac{1}{x-2}-\frac{12}{x^3-8})$$

Answered by prakash jain last updated on 24/Dec/17

$$\frac{1}{x-2}-\frac{1}{(x-2)(x^2+2x+4)}$$$$=\frac{x^2+2x+4-12}{(x-2)(x^2+2x+4)}$$$$=\frac{x^2+2x-8}{(x-2)(x^2+2x+4)}=\frac{(x-2)(x+4)}{(x-2)(x^2+2x+4)}$$$$\underset{x\to 2}{\lim }\frac{(x-2)(x+4)}{(x-2)(x^2+2x+4)}=\frac{2+4}{4+4+4}=\frac{1}{2}$$

Commented by Rasheed.Sindhi last updated on 24/Dec/17

$$\mathrm{pl}\mathrm{check}\mathrm{last}\mathrm{step}\frac{2+4}{4+4+4}=\frac{6}{12}=\frac{1}{2}$$

Commented by prakash jain last updated on 24/Dec/17

Thanks Corrected

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