Posted by
**Emily** on
.

uu) lim ln(1-x) as x->1-

yy) lim (√(6-x)-2)/(√(3-x)-1) as x-> 2

zz) lim (1-(1/2)arctanx) as x->-∞

bbb) lim ln|1-x| as x->1

If these are hard to interpret with all the parentheses, you can plug them in to Wolfram Alpha and it interprets them perfectly.