2
$\begingroup$

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

No idea where to even start on this one.

3 Answers 3

4

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

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

  • 0
    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.2011-08-28
  • 0
    @Jordan: I was hoping you would get $\dfrac{x+4}{4x(4+x)}$ which can be simplified further2011-08-28
  • 0
    Yes that is what I got m y second attempt and then simplified that into -1/162011-08-28
  • 3
    @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
0

Multiply everything by $4x$:

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

Cancel stuff out:

$\frac{1}{4x}$

Input $-4$:

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

0

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