lim(n2+2n+2)/(n+1)-an)=b,求a,bn2是n平方lim((n平方+2n+2)/(n+1)-an)=b
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![lim(n2+2n+2)/(n+1)-an)=b,求a,bn2是n平方lim((n平方+2n+2)/(n+1)-an)=b](/uploads/image/z/6458754-66-4.jpg?t=lim%EF%BC%88n2%2B2n%2B2%EF%BC%89%2F%EF%BC%88n%2B1%EF%BC%89-an%EF%BC%89%3Db%2C%E6%B1%82a%2Cbn2%E6%98%AFn%E5%B9%B3%E6%96%B9lim%EF%BC%88%EF%BC%88n%E5%B9%B3%E6%96%B9%2B2n%2B2%EF%BC%89%2F%EF%BC%88n%2B1%EF%BC%89-an%EF%BC%89%3Db)
lim(n2+2n+2)/(n+1)-an)=b,求a,bn2是n平方lim((n平方+2n+2)/(n+1)-an)=b
lim(n2+2n+2)/(n+1)-an)=b,求a,b
n2是n平方
lim((n平方+2n+2)/(n+1)-an)=b
lim(n2+2n+2)/(n+1)-an)=b,求a,bn2是n平方lim((n平方+2n+2)/(n+1)-an)=b
a=1,b=1
lim里面先化简,得到n+1+1/(n+1)-an,即lim[(1-a)n+1+1/(n+1)]=b
所以显然有a=1,b=1
没问题了吧