   here they are

reopened

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(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

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