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\(a=\lim\dfrac{5n\left(n+\sqrt{n^2-n-1}\right)}{n+1}=\lim\dfrac{5\left(n+\sqrt{n^2-n-1}\right)}{1+\dfrac{1}{n}}=\dfrac{+\infty}{1}=+\infty\)
\(b=\lim\dfrac{\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}+\dfrac{1}{n^4}}}}{1-\dfrac{1}{\sqrt{n}}}=\dfrac{0}{1}=0\)
\(c=\lim\dfrac{\sqrt{2n^2-1+\dfrac{7}{n^2}}}{3+\dfrac{5}{n}}=\dfrac{+\infty}{3}=+\infty\)
\(d=\lim\dfrac{\sqrt{3+\dfrac{2}{n}}-1}{3-\dfrac{2}{n}}=\dfrac{\sqrt{3}-1}{3}\)
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
\(lim\dfrac{5n\sqrt{2n^2-n}}{1+5n-3n^2}=lim\dfrac{5\sqrt{2-\dfrac{1}{n}}}{\dfrac{1}{n^2}+\dfrac{5}{n}-3}=\dfrac{5\sqrt{2-0}}{0+0-3}=\dfrac{-5\sqrt{2}}{3}\)
\(lim\dfrac{\sqrt{4n^2+n}-7n}{3n^2-1}=lim\dfrac{\sqrt{\dfrac{4}{n^2}+\dfrac{1}{n^3}}-\dfrac{7}{n}}{3-\dfrac{1}{n^2}}=\dfrac{\sqrt{0+0}-0}{3-0}=\dfrac{0}{3}=0\)
\(lim\left(5n-\sqrt{25n^2-3n+5}\right)=lim\dfrac{25n^2-25n^2+3n-5}{5n+\sqrt{25n^2-3n+5}}\)
\(=lim\dfrac{3n-5}{5n+\sqrt{25n^2-3n+5}}=lim\dfrac{3-\dfrac{5}{n}}{5+\sqrt{25-\dfrac{3}{n}+\dfrac{5}{n^2}}}=\dfrac{3-0}{5+\sqrt{25-0+0}}=\dfrac{3}{10}\)
\(lim\dfrac{4n^5-3n^4-2n^3+7n-9}{-5n\left(3n^2-3n+1\right)\left(5-2n^2\right)}=lim\dfrac{\dfrac{4n^5-3n^4-2n^3+7n-9}{n^5}}{\dfrac{-5n}{n}\dfrac{\left(3n^2-3n+1\right)}{n^2}\dfrac{\left(5-2n^2\right)}{n^2}}\)
\(=lim\dfrac{4-\dfrac{3}{n}-\dfrac{2}{n^2}+\dfrac{7}{n^4}-\dfrac{9}{n^5}}{-5.\left(3-\dfrac{2}{n}+\dfrac{1}{n^2}\right).\left(\dfrac{5}{n^2}-2\right)}=\dfrac{4-0-0+0-0}{-5\left(3-0+0\right).\left(0-2\right)}=\dfrac{2}{15}\)
\(=\lim\dfrac{\dfrac{1}{n}-3}{2+\dfrac{5}{n}-\dfrac{2}{n^3}}=-\dfrac{3}{2}\)
a,\(lim\dfrac{n^2-2n}{5n+3n^2}=lim\dfrac{1-\dfrac{2}{n}}{\dfrac{5}{n}+3}=\dfrac{1}{3}\)
b,\(lim\dfrac{n^2-2}{5n+3n^2}=lim\dfrac{1-\dfrac{2}{n^2}}{\dfrac{5}{n}+3}=\dfrac{1}{3}\)
c,\(lim\dfrac{1-2n}{5n+3n^2}=lim\dfrac{1-2n}{n\left(5+3n\right)}=lim\dfrac{\dfrac{1}{n}-2}{1\left(\dfrac{5}{n}+3\right)}=-\dfrac{2}{3}\)
d,\(lim\dfrac{1-2n^2}{5n+5}=lim\dfrac{\left(1-n\sqrt{2}\right)\left(1+n\sqrt{2}\right)}{5n+5}=lim\dfrac{\left(\dfrac{1}{n}-\sqrt{2}\right)\left(\dfrac{1}{n}+\sqrt{2}\right)}{5+\dfrac{5}{n}}=\dfrac{-2}{5}\)