(1) Let x=4+h, then we have (3-√(4+h+5))/h, (3-√(9+h))/h=(3-√(9(1+h/9)))/h,
(3-3√(1+h/9))/h=(3-3(1+h/9)^½)/h=(3-3(1+h/18))/h when h is small.
This becomes (3-3-h/6)/h=-⅙ as h→0, that is, in the limit as x→4.
(2) Factorise:
(x-3)(x+3)(x²+9)/((2x+1)(x-3))=(x+3)(x²+9)/(2x+1) when x≠3.
Therefore limit as x→3 is (3+3)(9+9)/(6+1)=6×18/7=108/7.
(3) √(3x-5)/5=√(30-5)/5=√25/5=5/5=1 as x→10, by directly putting x=10 in the expression, because there are potential anomalies here.
(4) Let x=1+h, then
((1+h)^⅓-1)/((1+h)^¼-1)=(1+h/3-1)/(1+h/4-1)=(h/3)/(h/4)=4/3 as h→0, x→1.
(5) (cos(2x)-1)/(cos(x)-1)=(2cos²(x)-1-1)/(cos(x)-1)=
2(cos²(x)-1)/(cos(x)-1)=2(cos(x)-1)(cos(x)+1)/(cos(x)-1)=
2(cos(x)+1) as x→0, which is 2(2)=4.
(6) No potential anomaly here, so just put x=3:
(45-24-13)/(9-5)=8/4=2