# Finding the limit of a quotient involving fractions

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

No idea where to even start on this one.

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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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That gives me (x+4)/(16x + 4x^2) which I do not think is correct. Actually I did it again and got -1/16 which I think is correct, not sure what went wrong first time. – user138246 Aug 28 '11 at 1:22
@Jordan: I was hoping you would get $\dfrac{x+4}{4x(4+x)}$ which can be simplified further – Henry Aug 28 '11 at 1:25
Yes that is what I got m y second attempt and then simplified that into -1/16 – user138246 Aug 28 '11 at 1:27
@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 $$\lim_{x\rightarrow -4}\frac{\frac{1}{4}+\frac{1}{x}}{4+x}=\lim_{x\rightarrow -4}\frac{1}{4x}=\frac{1}{-16}=-\frac{1}{16}.$$ – Américo Tavares Aug 28 '11 at 1:31

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 $-\frac{1}{16}$.

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