1.

lim(x_4) ∛(x/(-7x+1)) = ∛(4/(-7.4+1) =

2.

lim(x_2) (-2x^2+4x)/(x-2)

= lim(x_2) (-2x)(x-2)/(x-2)

= lim(x_2) (-2x) = -2.2 = -4

3.

lim(x_2) (x^2-16)/(x^2-5x-6)

= (2^2-16)/(2^2-5.2-6) = -12/-22 = 6/11

4.

lim(x_1/2) (3+2x)/(5-x)

= (3+2.(1/2)/(5-1/2) = 4/(9/2) = 8/9

5.

lim(x_3) (x^3-27)/(x-3)

= lim(x_3) [(x-3)(x^2+9) - (-3x^2+9x)]/(x-3)

= lim(x_3) [(x-3)(x^2+9) + 3x(x-3)]/(x-3)

= lim(x_3) (x^2+9) + 3x

= 3^2+9+3.3 = 27

6.

lim(y_-3) √[(y^2-y)/(2y^2+7y+3)]

= lim(y_-3) √[y(y-1)/(2y+1)(y+3)]

Not defined because,

lim(y_-3⁻) √[y(y-1)/(2y+1)(y+3)] = ∞

Going to -3 from left hand, its denominator have value almost positif zero. So, positif numerator divided by almost positif zero is going bigger and bigger to infinity and then sqrt of it is infinity.

lim(y_-3⁺) √[y(y-1)/(2y+1)(y+3)] = i ∞

Going to -3 from right hand, its denominator have value almost negative zero. So, positif numerator divided by almost negative zero is going bigger and bigger to negative infinity and then sqrt of it is i times infinity.

7.

lim(y_3) √[(2y^3-5y^2-2y-3)/(4y^3-13y^2+4y-3)]

= lim(y_3) √[(y-3)(2y^2+y+1)/(y-3)(4y^2-y+3)]

= lim(y_3) √[(2y^2+y+1)/(4y^2-y+3)]

= √[(2(3)^2+3+1)/(4(3)^2-3+3)] = √(22/36)

Untuk no. 8 sampai selanjutnya semoga yang lain bisa melengkapi