这道高数题是怎么做的,第二步看不懂
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consider
lim(x->0) (√[1+(sinx)^2] +1)/[ √(1+tanx) +√(1+sinx) ] =1
lim(x->0) { (√[1+(sinx)^2] +1)/[ √(1+tanx) +√(1+sinx) ] }. ( secx-1)/(xsinx)
=lim(x->0) ( secx-1)/(xsinx)
=lim(x->0) ( 1-cosx)/(xsinxcosx)
=lim(x->0) 2(sin(x/2))^2/(xsinxcosx)
lim(x->0) (√[1+(sinx)^2] +1)/[ √(1+tanx) +√(1+sinx) ] =1
lim(x->0) { (√[1+(sinx)^2] +1)/[ √(1+tanx) +√(1+sinx) ] }. ( secx-1)/(xsinx)
=lim(x->0) ( secx-1)/(xsinx)
=lim(x->0) ( 1-cosx)/(xsinxcosx)
=lim(x->0) 2(sin(x/2))^2/(xsinxcosx)
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