Очень срочно нужно помогите решить пож

Решение 1)  lim x---> 0 [sin(x/5) / 3x] = (1/15) * lim x---> 0 [sin(x/5) / (x/)5] =  1/15 2) lim x---> ∞ (1 + 1/3x) ^4x = lim x---> ∞ [(1 + 1/3x) ^3x]⁴/³ = e⁴/³ 3)  lim x---> 0 [sin(7x² + 5x - 11) / (7x² + 5x - 11)] = 1 4)   lim x---> π/4 [(cosx - sinx)/(tgx - 1)] =  =  lim x---> π/4 [(cosx - sinx)/(sinx/cosx - 1)] =  =  lim x---> π/4 [(cos²x - sinx*cosx)/(sinx - cosx)] =  = -  lim x---> π/4 [cosx*(cosx - sinx)/(cosx - sinx)] =   = -  lim x---> π/4(cosx) = - cos(π/4) = - √2/2