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1.
\(lim\frac{3n+1}{\sqrt[3]{\left(n^3+3n+1\right)^2}+n\sqrt{n^3+3n+1}+n^2}=lim\frac{3+\frac{1}{n}}{\sqrt[3]{\frac{\left(n^3+3n+1\right)^2}{n^3}}+\sqrt{n^3+3n+1}+n}=\frac{3}{\infty}=0\)
b=\(lim\left(\sqrt[3]{n^3+2n}-n+n-\sqrt{n^2+1}\right)=lim\left(\frac{2n}{\sqrt[3]{\left(n^3+2n\right)^2}+n\sqrt[3]{n^3+2n}+n^2}-\frac{1}{n+\sqrt{n^2+1}}\right)\)
\(=lim\left(\frac{2}{\sqrt[3]{\frac{\left(n^3+2n\right)^2}{n^3}}+\sqrt[3]{n^3+2n}+n}-\frac{1}{n+\sqrt{n^2+1}}\right)=0-0=0\)
c\(=lim\left(\frac{2n^2+n}{\sqrt[3]{\left(n^3+n\right)^2}+\sqrt[3]{\left(n^3+n\right)\left(n^3-2n^2\right)}+\sqrt[3]{\left(n^3-2n^2\right)^2}}\right)\)
\(=lim\left(\frac{2+\frac{1}{n}}{\sqrt[3]{\left(1+\frac{1}{n^2}\right)^2}+\sqrt[3]{\left(1+\frac{1}{n^2}\right)\left(1-\frac{2}{n}\right)}+\sqrt[3]{\left(1-\frac{2}{n}\right)^2}}\right)=\frac{2}{1+1.1+1}=\frac{2}{3}\)
2.
a\(=lim\left[n\left(2-\sqrt{1+\frac{3}{n}}\right)\right]=+\infty\left(2-1\right)=+\infty\)
\(b=lim\left[n\left(\sqrt{1+\frac{2}{n^2}}-\sqrt{\frac{3}{n}+\frac{1}{n^2}}\right)\right]=+\infty\left(1-0\right)=+\infty\)
\(c=lim\left[n^3\left(\frac{sin2n}{n^2}-3\right)\right]=+\infty\left(0-3\right)=-\infty\)
a/
\(lim\dfrac{\sqrt{n^2-n}-n}{n}=lim\dfrac{-n}{n\left(\sqrt{n^2-n}+n\right)}=lim\dfrac{-\dfrac{1}{n}}{1\left(\sqrt{1-\dfrac{1}{n}}+1\right)}=\dfrac{0}{2}=0\)
b/
\(lim\dfrac{2^n-5^{n+2}}{5^n-4^{2-n}}=lim\dfrac{8^n-25.20^n}{20^n-4^2}=lim\dfrac{\left(\dfrac{8}{20}\right)^n-25}{1-\dfrac{16}{20^n}}=\dfrac{0-25}{1-0}=-25\)
1) \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5n-3}{-n+5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(3n+5-\dfrac{3}{n}\right)}{-n\left(1-\dfrac{5}{n}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n+5-\dfrac{3}{n}}{-\left(1-\dfrac{5}{n}\right)}\)
\(=\left[{}\begin{matrix}-\infty\left(n\rightarrow+\infty\right)\\+\infty\left(n\rightarrow-\infty\right)\end{matrix}\right.\)
Bài 2,3 tương tự, bạn tự làm nhé!
1/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}.\sqrt[4]{1+8x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}-1}{x}\)
Liên hợp dài quá ko muốn gõ tiếp, bạn tự đặt nhân tử chung rồi liên hợp nhé, kết quả ra 5
2/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{1+7x}-2-\left(x^3-3x+2\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{7\left(x-1\right)}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)^2\left(x+2\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{7}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)\left(x+2\right)=\dfrac{7}{12}\)
3/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x^3-x^2+1}{2x^2+3x-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{x-1+\dfrac{1}{x^2}}{2+\dfrac{3}{x}-\dfrac{1}{x^2}}=-\infty\)
4/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}}{\sqrt{4x+1}}=\lim\limits_{x\rightarrow+\infty}\dfrac{1+\dfrac{1}{\sqrt[6]{x}}+\dfrac{1}{\sqrt[4]{x}}}{\sqrt{4+\dfrac{1}{x}}}=\dfrac{1}{\sqrt{4}}=\dfrac{1}{2}\)
5/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+\sqrt{x^2+2}}{\sqrt[3]{8x^3+x^2+1}}=\lim\limits_{x\rightarrow-\infty}\dfrac{1-\sqrt{1+\dfrac{2}{x^2}}}{\sqrt[3]{8+\dfrac{1}{x}+\dfrac{1}{x^3}}}=\dfrac{1-1}{\sqrt[3]{8}}=0\)
6/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+3x-7}}{\sqrt[3]{27x^3+5x^2+x-4}}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4+\dfrac{3}{x}-\dfrac{7}{x^2}}}{\sqrt[3]{27+\dfrac{5}{x}+\dfrac{1}{x^2}-\dfrac{4}{x^3}}}=\dfrac{-\sqrt{4}}{\sqrt[3]{27}}=\dfrac{-2}{3}\)
a)
=
= -4.
b)
=
=
(2-x) = 4.
c)
=
=
=
=
.
d)
=
= -2.
e)
= 0 vì
(x2 + 1) =
x2( 1 +
) = +∞.
f)
=
= -∞, vì
> 0 với ∀x>0.
a) 
(x4 – x2 + x - 1) = x4(1 - 
) = +∞.
b) 
(-2x3 + 3x2 -5 ) = x3(-2 + 
) = +∞.
c) 
= 
= +∞.
d) \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+1}+x}{5-2x}=\lim\limits_{x\rightarrow+\infty}\dfrac{\left|x\right|\sqrt{1+\dfrac{1}{x^2}}+x}{5-2x}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{1+\dfrac{1}{x^2}}+x}{5-2x}\)\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x^2}}+1}{\dfrac{5}{x}-2}=-1\).
1: \(\lim\limits_{n\rightarrow\infty}\dfrac{-7n^2+4}{-n+5}=\lim\limits_{n\rightarrow\infty}\dfrac{7n^2-4}{n-5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(7-\dfrac{4}{n^2}\right)}{n\left(1-\dfrac{5}{n}\right)}=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(7-\dfrac{4}{n^2}\right)}{1-\dfrac{5}{n}}\)
\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n=+\infty\\\lim\limits_{n\rightarrow\infty}\dfrac{7-\dfrac{4}{n^2}}{1-\dfrac{5}{n}}=\dfrac{7}{1}=7>0\end{matrix}\right.\)
2:
\(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^2+2}{n-2}=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(-3+\dfrac{2}{n^2}\right)}{n\left(1-\dfrac{2}{n}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(-3+\dfrac{2}{n^2}\right)}{1-\dfrac{2}{n}}\)
\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n=+\infty\\\lim\limits_{n\rightarrow\infty}\dfrac{-3+\dfrac{2}{n^2}}{1-\dfrac{2}{n}}=-\dfrac{3}{1}=-3< 0\end{matrix}\right.\)