# What is the limit of (1 / (4x - 4x^2) ) - (1 / (4x - 5x^2) ) as x approaches 0?

Apr 14, 2016

$- \frac{1}{16}$

#### Explanation:

You wish to find

${\lim}_{x \rightarrow 0} \left(\frac{1}{4 x - 4 {x}^{2}} - \frac{1}{4 x - 5 {x}^{2}}\right)$

Note first that plugging in $0$ for $x$ will cause fractions with denominators of $0$, so we can't determine the limit without doing some manipulation.

First, factor the fractions' denominators:

$= {\lim}_{x \rightarrow 0} \left(\frac{1}{x \left(4 - 4 x\right)} - \frac{1}{x \left(4 - 5 x\right)}\right)$

Now, find a common denominator of $x \left(4 - 4 x\right) \left(4 - 5 x\right)$.

$= {\lim}_{x \rightarrow 0} \left(\frac{4 - 5 x}{x \left(4 - 4 x\right) \left(4 - 5 x\right)} - \frac{4 - 4 x}{x \left(4 - 4 x\right) \left(4 - 5 x\right)}\right)$

Combine the fraction.

$= {\lim}_{x \rightarrow 0} \frac{\left(4 - 5 x\right) - \left(4 - 4 x\right)}{x \left(4 - 4 x\right) \left(4 - 5 x\right)}$

Simplify the numerator.

$= {\lim}_{x \rightarrow 0} \frac{4 - 5 x - 4 + 4 x}{x \left(4 - 4 x\right) \left(4 - 5 x\right)}$

$= {\lim}_{x \rightarrow 0} \frac{- x}{x \left(4 - 4 x\right) \left(4 - 5 x\right)}$

Cancel the $x$ term in the numerator and denominator.

$= {\lim}_{x \rightarrow 0} \frac{- 1}{\left(4 - 4 x\right) \left(4 - 5 x\right)}$

We can now evaluate the limit by plugging in $0$ for $x$.

$= \frac{- 1}{\left(4 - 4 \left(0\right)\right) \left(4 - 5 \left(0\right)\right)} = \frac{- 1}{\left(4 - 0\right) \left(4 - 0\right)} = - \frac{1}{16}$

If we graph the original function, we should see that the graph approaches $- \frac{1}{16}$ at $x = 0$, even though the function is undefined at $x = 0$:

graph{1/(4x-4x^2)-1/(4x-5x^2) [-1.662, 2.664, -1.718, 0.445]}