Question

第18题怎么做 求解答

Answer 1

= （x→0）lim { [ln(1+x²)] / [ -cost|（0至x）] }
= （x→0）lim { [ln(1+x²)] / [ -cosx+cos0 ] }
= （x→0）lim { [ln(1+x²)] / [ 1-cosx ] }
= （x→0）lim { [2x/(1+x²)] / sinx }
= （x→0）lim { [2x/(1+x²)] /x }
= （x→0）lim {2/(1+x²) }
= 2/(1+0)
= 2