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Calculate #lim_(x ->x_o) (x^2- x_o x)/(x^2-x_o^2)# for every value of #x_o in RR# ?

2 Answers
Feb 18, 2018

Answer:

#lim_(x→x_0) (x^2-x_0x)/(x^2-x_0^2) = 1/2#

Explanation:

#lim_(x→x_0) (x^2-x_0x)/(x^2-x_0^2) = lim_(x→x_0) (x(x-x_0))/((x-x_0)(x+x_0))=lim_(x→x_0) x/(x+x_0)#

We can now let #x=x_0# to get the limit as

#x_0/(x_0+x_0) = x_0/(2x_0)=1/2#

Feb 18, 2018

Answer:

#lim_(x->x_0) (x^2-x_0x)/(x^2-x_0^2) = {(1/2 " for " x_0 !=0), (1 " for " x_0 = 0):}#

Explanation:

Simplify the function:

#(x^2-x_0x)/(x^2-x_0^2) = (x(x-x_0))/((x+x_0)(x-x_0))=x/(x+x_0)#

so for #x_0 !=0#:

#lim_(x->x_0) (x^2-x_0x)/(x^2-x_0^2) = lim_(x->x_0) x/(x+x_0) = x_0/(x_0+x_0) = 1/2#

while for #x_0 = 0#:

#lim_(x->x_0) (x^2-x_0x)/(x^2-x_0^2) = lim_(x->0) x^2/x^2 = 1#