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

1 Answer
Apr 14, 2016

#-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]}