Question #7c54f

#Lim_(xrarr0)sqrt(x²+2x+2)-x=sqrt2#

or

#Lim_(xrarroo)sqrt(x²+2x+2)-x*(sqrt(x²+2x+2)+x)/(sqrt(x²+2x+2)+x)#

#Lim_(xrarroo)(x²+2x+2-x^2)/(sqrt(x²+2x+2)+x)#

#Lim_(xrarroo)(2x+2)/(sqrt(x²+2x+2)+x)#

#Lim_(xrarroo)(cancelx(2+2/x))/(cancelx(sqrt(1+2/x+2/x^2)+1)#

#Lim_(xrarroo)(2+2/x)/(sqrt(1+2/x+2/x^2)+1)=2/(1+1)=1#

You have a mistake but a correct solution :D

#((sqrt(x²+2x+2)-x))/1 * ((sqrt(x²+2x+2)+x))/((sqrt(x²+2x+2)+x)) = ((x^2+2x+2)-x^2)/(sqrt(x²+2x+2)+x) #

# = (2x+2)/(sqrt(x^2(1+2/x+2/x^2))+x)# #" "# for #x != 0#

# = (2x+2)/(sqrt(x^2)sqrt(1+2/x+2/x^2)+x)#

Use #sqrt(x^2) = absx# and #x > 0# (so #absx=x#) to get

# = (2x+2)/(xsqrt(1+2/x+2/x^2)+x)#

# = (x(2+2/x))/(x(sqrt(1+2/x+2/x^2)+1))#

# = (2+2/x)/(sqrt(1+2/x+2/x^2)+1)#

Now find the limit (I assume we are finding the limit at infinity)

#lim_(xrarroo)(sqrt(x²+2x+2)-x) = lim_(xrarroo)(2+2/x)/(sqrt(1+2/x+2/x^2)+1)#

# = (2+0)/(sqrt(1+0+0)+1) = 2/2=1#

Note that
Your #2x^2# in the numerator is incorrect

and

#lim_(xrarr00)(2x+2+(2/x))/(sqrt(1+(2/x)+(2/x²))+1) = (oo+2+0)/(sqrt1+1) = oo# #" "#

The limit of your expression is not #" "# #2/2=1#