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$\lim_{x\to -4} (1/4 + 1/x)/(4+x)$

No idea where to even start on this one.

3 Answers 3

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Hint: how about simplifying this by multiplying numerator and denominator by $4x$?

Or try to simplify $\dfrac{1}{4(4+x)} + \dfrac{1}{x(4+x)}$?

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    @Jordan Carlyon: $-1/16$ is correct. Since $\frac{\frac{1}{4}+\frac{1}{x}}{4+x}=\frac{\frac{x+4}{4x}}{4+x}=\frac{x+4}{4x}\times\frac{1}{4+x}=\frac{1}{4x},$ then $\lim_{x\rightarrow -4}\frac{\frac{1}{4}+\frac{1}{x}}{4+x}=\lim_{x\rightarrow -4}\frac{1}{4x}=\frac{1}{-16}=-\frac{1}{16}.$2011-08-28
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Multiply everything by $4x$:

$\frac{x+4}{4x(4+x)}$

Cancel stuff out:

$\frac{1}{4x}$

Input $-4$:

$\frac{1}{4(-4)} = -\frac{1}{16}$

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I would guess that the intention of the question is for you to notice that you can just add the fractions in the numerator in order to get cancelling terms. So the numerator becomes:

$ \dfrac{1}{4} + \dfrac{1}{x} = \dfrac{4 + x}{4x}$

And so the original expression becomes:

$ \dfrac{\frac{4+x}{4x}}{4+x} = \dfrac{1}{4x}$

Obviously from here the limit can be seen to be $-\frac{1}{16}$.