How do you find the limit of #sqrt(1- 8x^3) /sqrt(1-4x^2)# as x approaches 1/2?

We will use the following identities, which are a difference of cubes and a difference of squares.

• #1-8x^3=1^3-(2x)^3=(1-2x)(1+2x+4x^2)#
• #1-4x^2=1-(2x)^2=(1+2x)(1-2x)#

Thus, we have the limit

#lim_(xrarr1/2)sqrt(1-8x^3)/sqrt(1-4x^2)=lim_(xrarr1/2)sqrt((1-2x)(1+2x+4x^2))/sqrt((1+2x)(1-2x))#

Square roots can be split up through multiplication:

#=lim_(xrarr1/2)(sqrt(1-2x)sqrt(1+2x+4x^2))/(sqrt(1-2x)sqrt(1+2x))#

Cancel the #sqrt(1-2x)# terms:

#=lim_(xrarr1/2)sqrt(1+2x+4x^2)/sqrt(1+2x)#

We can now evaluate the limit by plugging in #1/2# for #x#:

#=sqrt(1+2(1/2)+4(1/2)^2)/sqrt(1+2(1/2))=sqrt(1+1+1)/sqrt(1+1)=sqrt(3/2)#

This can also be written as #sqrt6/2#.

Note that #sqrt6/2approx1.225# and examine the graph of the function:

graph{sqrt(1-8x^3)/sqrt(1-4x^2) [-2.827, 2.648, -0.39, 2.347]}

The graph of the function appears to be approaching #approx1.225# at #x=1/2#.

Note:

Upon examining the graph, it becomes clear that this is a one-sided limit (from the left). This occurs since for #x>1/2#, we have a negative value inside the square root.

Thus, the accurate statement to make is that the limit from the left approaches #sqrt6/2#, expressed mathematically as:

#lim_(xrarr1/2color(white)( )^-)sqrt(1-8x^3)/sqrt(1-4x^2)=sqrt6/2#