# How do you find the limit of ((1/x+16)-(1/16)) / x as x approaches 0?

May 22, 2015

I think you meant: $\frac{\frac{1}{x + 16} - \frac{1}{16}}{x}$.

When we try to find the limit, we get the (indeterminate) form $\frac{0}{0}$. We need to rewrite and try again.

The important algebra can be done in a couple ways:

Method 1

"If I had a fraction over a fraction, I'd know what to do next."

Good! Make it so.

$\frac{\frac{1}{x + 16} - \frac{1}{16}}{x} = \frac{\frac{16}{16 \left(x + 16\right)} - \frac{x + 16}{16 \left(x + 16\right)}}{\frac{x}{1}}$

$= \frac{\frac{16 - \left(x + 16\right)}{16 \left(x + 16\right)}}{\frac{x}{1}}$

$= \frac{- x}{16 \left(x + 16\right)} \cdot \frac{1}{x}$

$= \frac{- 1}{16 \left(x + 16\right)}$

Method 2

"I know this trick:"
Multiply numerator and denominator by the common denominator of all the fractions in the numerator and denominator. (Sounds complicated, but it's quicker.)

The common denominator is $16 \left(x + 16\right)$, so:

(1/(x+16) - 1/16)/x = ((1/(x+16) - 1/16))/x (16(x+16))/(16(x+16)

= ((16(x+16))/(x+16) - (16(x+16))/16)/(x(16(x+16))

$= \frac{16 - \left(x + 16\right)}{16 x \left(x + 16\right)}$

$= \frac{- 1}{16 \left(x + 16\right)}$

So, we get:

${\lim}_{x \rightarrow 0} \frac{\frac{1}{x + 16} - \frac{1}{16}}{x} = {\lim}_{x \rightarrow 0} \frac{- 1}{16 \left(x + 16\right)} = \frac{- 1}{16} ^ 2 = \frac{- 1}{256}$.