Find the limit: $$\lim_{x \to -{6}} \frac{(1/6)+(1/x)}{6+x}$$
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If you let $g(x)=\frac{1}{x}$ then this can be written as:$$\lim_{x\to-6} \frac{g(x)-g(-6)}{x-(-6)}$$ So the value is $g'(-6)$. – Thomas Andrews Apr 18 '14 at 16:17
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$$\frac{\dfrac16+\dfrac1x}{6+x}=\frac{6+x}{(6+x)6x}=\frac1{6x}$$ if $x+6\ne0$
If $x\to-6,x\ne-6$
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\begin{equation} \require{cancel} \lim_{x\to-6}\frac{\cfrac{1}{6}+\cfrac{1}{x}}{6+x}=\lim_{x\to-6}\frac{\cancel{(x+6)}}{6x\cancel{(x+6)}}=\lim_{x\to-6}\frac1{6x}=-\frac{1}{36}. \end{equation}
Anastasiya-Romanova 秀
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