maths

1)
Let 1/m = M and 1/x = X →∞ as x→0 , thenlim (1 + x/m)¹ˡ ˣ
x→0= lim (1 + M/X)ˣ
X→∞= eᴹ= e¹ˡ ᵐ

2)For the line y = 4x + 8 , the gradient = 4.
For the curve C : y = (x - 1)⁴+ 4 , the gradient dy/dx = 4(x - 1)³.
∴ 4(x - 1)³ = 4
⇒ x = 2Sub. into y = (x - 1)⁴+ 4 ,
y = 5.The required equation is (y - 5) / (x - 2) = 4 ,
i.e. y = 4x - 3

3)√(a + b√3) = 13 / (4 + √3)
√(a + b√3) = 13(4 - √3) / [(4 + √3)(4 - √3)]
√(a + b√3) = 4 - √3
√(a + b√3) = √(4 - √3)²
√(a + b√3) = √(19 - 8√3)
∴ a = 19 , b = - 8