Question

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

Answer 1

`-1/16`

Explanation:

You wish to find

`lim_(xrarr0)(1/(4x-4x^2)-1/(4x-5x^2))`

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_(xrarr0)(1/(x(4-4x))-1/(x(4-5x)))`

Now, find a common denominator of `x(4-4x)(4-5x)`.

`=lim_(xrarr0)((4-5x)/(x(4-4x)(4-5x))-(4-4x)/(x(4-4x)(4-5x)))`

Combine the fraction.

`=lim_(xrarr0)((4-5x)-(4-4x))/(x(4-4x)(4-5x))`

Simplify the numerator.

`=lim_(xrarr0)(4-5x-4+4x)/(x(4-4x)(4-5x))`

`=lim_(xrarr0)(-x)/(x(4-4x)(4-5x))`

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

`=lim_(xrarr0)(-1)/((4-4x)(4-5x))`

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

`=(-1)/((4-4(0))(4-5(0)))=(-1)/((4-0)(4-0))=-1/16`

If we graph the original function, we should see that the graph approaches `-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]}