#lim_(x - 0) 1/x (2/(2+x)^3-1/4) = # ?

Question

#lim_(x -> 0) 1/x (2/(2+x)^3-1/4) = # ?

Ezra Adams · Accepted Answer

#-3/8# Calling #f(x) = 2/(2+x)^3# we have #lim_(h->0)(f(0+h)-f(0))/h = f'(0)=-6/2^4 = -3/8#

Elijah Abram · Answer

Distribute the #1/x#
Make a common denominator
Simplify
Evaluate at 0

Given: #lim_(x to 0) 1/x (2/(2+x)^3-1/4)# Distribute the #1/x#: #lim_(x to 0) (2/(x(2+x)^3)-1/(4x))# We are aware that there are clever ways to multiply each fraction by one to create a common denominator: #lim_(x to 0) (2/(x(2+x)^3)4/4-1/(4x)(2+x)^3/(2+x)^3)# Execute the multiplication: #lim_(x to 0) (8/(4x(2+x)^3)-(2+x)^3/(4x(2+x)^3))# Over the common denominator, add the numerators: #lim_(x to 0) (8-(2+x)^3)/(4x(2+x)^3)# In the numerator, enlarge the cube: #lim_(x to 0) (8-(x^3 + 6 x^2 + 12 x + 8))/(4x(2+x)^3)# Assign the -1: #lim_(x to 0) (8-x^3 - 6 x^2 - 12 x - 8)/(4x(2+x)^3)# In the numerator, the #8-8 = 0#: #lim_(x to 0) (-x^3 - 6 x^2 - 12 x)/(4x(2+x)^3)# Removing a common factor of x is possible: #lim_(x to 0) (-x^2 - 6 x - 12)/(4(2+x)^3)# Now we can let #x to 0#: #(-0^2-6(0)-12)/(4(2+0)^3) = -3/8#

#lim_(x -&gt; 0) 1/x (2/(2+x)^3-1/4) = # ?

#lim_(x -> 0) 1/x (2/(2+x)^3-1/4) = # ?