Question

Решить пределы(подробно):
1) lim(x->0) 1-cos4x/6x^2
2) lim(x->∞) (3x+2 / 3x+5) ^x+2

Answer 1

$lim_{x to 0} frac{1-cos4x}{6x^2} =[ frac{0}{0} ]=lim_{x to 0} frac{4sin4x}{12x}=lim_{x to 0} frac{16x}{12x}= frac{16}{12} = frac{4}{3} =1 frac{1}{3} \ lim_{x to infty} ((1+ frac{-3}{3x+5} )^{ frac{3x+5}{-3} })^{ frac{-3(x+2)}{3x+5} }=e^{ lim_{x to infty} frac{-3(x+2)}{3x+5}}=e^{-1}$