How do you find the limit as x approaches infinity of a more complex square root function?
#lim_"x->oo"# #sqrt(x^2-4x+3)-sqrt(x^2+2x)#
1 Answer
The initial form is indeterminate (
Explanation:
# = ((x^2-4x+3)-(x^2+2x))/(sqrt(x^2-4x+3)+sqrt(x^2+2x))#
# = (-6x+3)/(sqrt(x^2-4x+3)+sqrt(x^2+2x))#
For
# = (-6x+3)/(sqrt(x^2)sqrt(1-4/x+3/x^2)+sqrt(x^2)sqrt(1+2/x))#
Now, since
# = (-6x+3)/(xsqrt(1-4/x+3/x^2)+xsqrt(1+2/x))# (for#x > 0# )
Now factor out the
# = (cancel(x)(-6+3/x))/(cancel(x)(sqrt(1-4/x+3/x^2)+sqrt(1+2/x)))# (for#x > 0# ).
Taking the limit as