lim( √n^2+2n)-n-1 lim( √4n^2+n)-n lim( √n^2-n)-n lim1/( √n^2+2) – ( √n^2+4)

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1. Đáp án:

    

   Giải thích các bước giải:

   lim(√n2+2n−n−1)=limn2+2n−(n+1)2√n2+2n+n+1=lim−1√n2+2n+n+1=lim−1n√1+2n+1+1n=0lim(√4n2+n−n)=lim4n2+n−n2√4n2+n+n=lim3+1n√4n2+1n4+1n=+∞lim(√n2−n−n)=limn2−n−n2√n2−n+n=lim−1√1−1n+1=−1lim(n2+2n−n−1)=limn2+2n−(n+1)2n2+2n+n+1=lim−1n2+2n+n+1=lim−1n1+2n+1+1n=0lim(4n2+n−n)=lim4n2+n−n24n2+n+n=lim3+1n4n2+1n4+1n=+∞lim(n2−n−n)=limn2−n−n2n2−n+n=lim−11−1n+1=−1

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2. Đáp án:

   $\begin{array}{l}
   \lim \left( {\sqrt {{n^2} + 2n}  – n – 1} \right)\\
    = \lim \frac{{{n^2} + 2n – {{\left( {n + 1} \right)}^2}}}{{\sqrt {{n^2} + 2n}  + n + 1}}\\
    = \lim \frac{{ – 1}}{{\sqrt {{n^2} + 2n}  + n + 1}}\\
    = \lim \frac{{ – \frac{1}{n}}}{{\sqrt {1 + \frac{2}{n}}  + 1 + \frac{1}{n}}} = 0\\
   \lim \left( {\sqrt {4{n^2} + n}  – n} \right)\\
    = \lim \frac{{4{n^2} + n – {n^2}}}{{\sqrt {4{n^2} + n}  + n}}\\
    = \lim \frac{{3 + \frac{1}{n}}}{{\sqrt {\frac{4}{{{n^2}}} + \frac{1}{{{n^4}}}}  + \frac{1}{n}}} =  + \infty \\
   \lim \left( {\sqrt {{n^2} – n}  – n} \right)\\
    = \lim \frac{{{n^2} – n – {n^2}}}{{\sqrt {{n^2} – n}  + n}}\\
    = \lim \frac{{ – 1}}{{\sqrt {1 – \frac{1}{n}}  + 1}} =  – 1
   \end{array}$

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