For the first limit, we will consider the Left and Right Handed
Limits of the Function : #f(x)=|10-2x|/(3x^2-13x-10)#.
#:. f(x)=|10-2x|/{(x-5)(3x+2)}, xne5,-2/3#.
Now, as #x to 5+, x >5, so, (x-5) >0 :. |x-5|=x-5#.
#rArr|10-2x|=|2(5-x)|=2|5-x|=2|x-5|=2(x-5)#.
#:. lim_(xto 5+)f(x)=lim_(x to 5+){2(x-5)}/{(x-5)(3x+2)}#,
#=lim_(xto 5+)2/(3x+2)=2/(3xx5+2)#.
# rArr lim_(x to 5+)f(x)=2/17................................(ast_1)#.
Similarly, for the #lim_(x to 5-)f(x)," since, "x <5; |x-5|=5-x#,
#rArr lim_(x to 5-)f(x)=-2/17..................................(ast_2)#.
#(ast_1) and (ast_2)# force to conclude that,
#lim_(x to 5)|10-2x|/{(x-5)(3x+2)}# does not exist.
For the second limit, we subst. #x=y^6#
Note that, #6,# being the lcm of #2 and 3#, it will help us to
get rid of both a square root (power #1/2#) and a cube root
(power #1/3#).
# x=y^6, and, x to 1 rArr y to 1#.
#:."The Reqd. Lim.="lim_(x to 1)(sqrtx-1)/(root(3)x-1)#,
#=lim_(y to 1)(sqrt(y^6)-1)/(root(3)(y^6)-1)#,
#=lim_(y to 1)(y^3-1)/(y^2-1)#,
#=lim_(y to 1){cancel((y-1))(y^2+y+1)}/{cancel((y-1))(y+1)}#,
#=lim_(y to 1)(y^2+y+1)/(y+1)#,
#=(1^2+1+1)/(1+1)#.
#rArr "The Reqd. Lim.="3/2#.
This limit can also be obtained using the following standard form :
#lim_(x to a)(x^n-a^n)/(x-a)=na^(n-1)#.
#lim_(x to 1)(sqrtx-1)/(root(3)x-1)#,
#=lim_(x to 1)(x^(1/2)-1^(1/2))/(x-1)-:(x^(1/3)-1^(1/3))/(x-1)#,
#=1/2*(1)^(1/2-1)-:1/3*(1)^(1/3-1)#,
#=3/2#, as before!
