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

Jul 1, 2016

left-sided limit is $\infty$

right-sided limit is $- \infty$

#### Explanation:

${\lim}_{x \to - 4} \frac{\frac{1}{4} x + \frac{1}{x}}{4 + x}$ and clearly that is $\frac{- \frac{5}{4}}{0}$ 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

${\lim}_{h \to 0} \frac{\frac{1}{4} \left(- 4 + h\right) + \frac{1}{- 4 + h}}{4 - 4 + h}$

${\lim}_{h \to 0} \frac{- 1 + \frac{h}{4} + \frac{1}{- 4 + h}}{h}$

we can expand this part of the numerator ie

$\frac{1}{- 4 + h} = \frac{- \frac{1}{4}}{1 - \frac{1}{4} h} = - \frac{1}{4} {\left(1 - \frac{h}{4}\right)}^{- 1}$

The binomial expansion is

$- \frac{1}{4} \left(1 - \left(- 1\right) \frac{h}{4} + m a t h c a l \left\{O\right\} \left({h}^{2}\right)\right)$

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

so the limit is now

${\lim}_{h \to 0} \frac{- 1 + \frac{h}{4} - \frac{1}{4} - \frac{h}{16} + m a t h c a l \left\{O\right\} \left({h}^{2}\right)}{h}$

${\lim}_{h \to 0} \frac{\textcolor{red}{- \frac{5}{4}} - \frac{3 h}{16} + m a t h c a l \left\{O\right\} \left({h}^{2}\right)}{\textcolor{red}{h}}$

IOW:

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

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