How do you find the limit of $\frac{\frac{1}{4} x + \frac{1}{x}}{4 + x}$ $(1/4x + 1/x) / (4+x)$ as x approaches -4?

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Answer 1 · 2016-07-01T17:40:07

left-sided limit is $\infty$ $oo$

right-sided limit is $- \infty$ $-oo$

$lim_{x to -4} (1/4x + 1/x) / (4+x)$ and clearly that is $(-5/4)/0$ at x = -4

we can let $x= -4 + h$ , so $h = x+4$ , where $0<|h| < < 1$ . This allows us to explore the effect of small variations $h$ about $x = -4$

so the limit becomes

$lim_{h to 0} (1/4(-4+h) + 1/(-4+h)) / (4-4 + h)$

$lim_{h to 0} (-1 + h/4 + 1/(-4+h)) / (h)$

we can expand this part of the numerator ie

$1/(-4+h) = (-1/4)/(1 - 1/4h) = - 1/4 (1 - h/4)^{-1}$

The binomial expansion is

$-1/4 (1 - (-1) h/4 + mathcal{O} (h^2))$

$-1/4 - h/16 + mathcal{O} (h^2))$

so the limit is now

$lim_{h to 0} (-1 + h/4 -1/4 - h/16 + mathcal{O} (h^2)) / (h)$

$lim_{h to 0} ( color{red}{-5/4} - (3h)/16 + mathcal{O} (h^2)) / (color{red}{h})$

IOW:

if h < 0, ie we are to the left of x = -4, the limit is $oo$

if h > 0, ie we are to right of x = -4, the limit is $-oo$

How do you find the limit of (1/4x + 1/x) / (4+x)14x+1x4+x (1/4x + 1/x) / (4+x) as x approaches -4?