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xlog (1 +x)limx0 (1-cos x) |
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Answer» if we put x=0 we will get 0/0 form so using L'hospital rule lim x->0 (log(1+x)+x/(1+x))/(sinx) if we put x=0 we will get 0/0 form so using L'hospital rule lim x->0 (1/(1+x)+1/(1+x)-x/(1+x)(1+x))/cosx = (1+1)/1 = 2 |
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