Question

очень срочно нужно помогите решить пож

Answer 1

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