`a)lim_{x->+oo}[x+1]/[x^2+x+1]`

`=lim_{x->+oo}[1/x+1/[x^2]]/[1+1/x+1/[x^2]]`

`=0`

`b)lim_{x->+oo}[3x+1]/[3x^2-x+5]`

`=lim_{x->+oo}[3/x+1/[x^2]]/[3-1/x+5/[x^2]]`

`=0`

`c)lim_{x->-oo}[3x+5]/[\sqrt{x^2+x}]`

`=lim_{x->-oo}[3+5/x]/[-\sqrt{1+1/x}]`

`=-3`

`d)lim_{x->+oo}[-5x+1]/[\sqrt{3x^2+1}]`

`=lim_{x->+oo}[-5+1/x]/[\sqrt{3+1/[x^2]}]`

`=-5/3`

a: \(=lim_{x->-\infty}\dfrac{2x-5+\dfrac{1}{x^2}}{7-\dfrac{1}{x}+\dfrac{4}{x^2}}\)

\(=\dfrac{2x-5}{7}\)

\(=\dfrac{2}{7}x-\dfrac{5}{7}\)

\(=-\infty\)

b: \(=lim_{x->+\infty}x\sqrt{\dfrac{1+\dfrac{1}{x}+\dfrac{3}{x^2}}{3x^2+4-\dfrac{5}{x^2}}}\)

\(=lim_{x->+\infty}x\sqrt{\dfrac{1}{3x^2+4}}=+\infty\)

a: \(\lim\limits_{x->0^-^-}\dfrac{-2x+x}{x\left(x-1\right)}=lim_{x->0^-}\left(\dfrac{-x}{x\left(x-1\right)}\right)\)

\(=lim_{x->0^-}\left(\dfrac{-1}{x-1}\right)=\dfrac{-1}{0-1}=\dfrac{-1}{-1}=1\)

b: \(=lim_{x->-\infty}\left(\dfrac{x^2-x-x^2+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)

\(=lim_{x->-\infty}\left(\dfrac{-x+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)

\(=lim_{x->-\infty}\left(\dfrac{-1+\dfrac{1}{x}}{-\sqrt{1-\dfrac{1}{x^2}}-\sqrt{1-\dfrac{1}{x^2}}}\right)=\dfrac{-1}{-2}=\dfrac{1}{2}\)

1.

Trước hết bạn nhớ công thức:

$1^2+2^2+....+n^2=\frac{n(n+1)(2n+1)}{6}$ (cách cm ở đây: https://hoc24.vn/cau-hoi/tinh-tongs-122232n2.83618073020)

Áp vào bài:

\(\lim\frac{1}{n^3}[1^2+2^2+....+(n-1)^2]=\lim \frac{1}{n^3}.\frac{(n-1)n(2n-1)}{6}=\lim \frac{n(n-1)(2n-1)}{6n^3}\)

\(=\lim \frac{(n-1)(2n-1)}{6n^2}=\lim (\frac{n-1}{n}.\frac{2n-1}{6n})=\lim (1-\frac{1}{n})(\frac{1}{3}-\frac{1}{6n})\)

\(=1.\frac{1}{3}=\frac{1}{3}\)

2.

\(\lim \frac{1}{n}\left[(x+\frac{a}{n})+(x+\frac{2a}{n})+...+(x.\frac{(n-1)a}{n}\right]\)

\(=\lim \frac{1}{n}\left[\underbrace{(x+x+...+x)}_{n-1}+\frac{a(1+2+...+n-1)}{n} \right]\)

\(=\lim \frac{1}{n}[(n-1)x+a(n-1)]=\lim \frac{n-1}{n}(x+a)=\lim (1-\frac{1}{n})(x+a)\)

\(=x+a\) 

3.

Trước tiên ta có công thức:

$1^3+2^3+....+n^3=(1+2+3+...+n)^2=\frac{n^2(n+1)^2}{4}$
Chứng minh: https://diendantoanhoc.org/topic/81694-t%C3%ADnh-t%E1%BB%95ng-s-13-23-33-n3/

Khi đó:

\(\lim \frac{1^3+2^3+...+n^3}{n^4}=\lim \frac{n^2(n+1)^2}{4n^4}\\ =\lim \frac{(n+1)^2}{4n^2}=\frac{1}{4}\lim (1+\frac{1}{n})^2=\frac{1}{4}.1=\frac{1}{4}\)

Sao anh không thấy đề cụ thể ta!

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+2x-1}-x}{3x-2}=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{\dfrac{4x^2+2x-1}{x^2}}-\dfrac{x}{x}}{\dfrac{3x-2}{x}}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4+\dfrac{2}{x}-\dfrac{1}{x^2}}-1}{3-\dfrac{2}{x}}=-\dfrac{4-1}{3}=-1\)

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+2x-1}-x}{3x-2}=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4+\dfrac{2}{x}-\dfrac{1}{x^2}}+1}{-3+\dfrac{2}{x}}=\dfrac{\sqrt{4}+1}{-3}=-1\).

a) = = -4.

b) = = (2-x) = 4.

c) =
= = = .

d) = = -2.

e) = 0 vì (x² + 1) = x²( 1 + ) = +∞.

f) = = -∞, vì > 0 với ∀x>0.

a: \(\lim\limits_{x\rightarrow+\infty}\dfrac{x-2}{3-\sqrt{x^2+7}}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1-\dfrac{2}{x}}{\dfrac{3}{x}-\sqrt{1+\dfrac{7}{x^2}}}\)

\(=\dfrac{1}{0-\sqrt{1+0}}=\dfrac{1}{-1}=-1\)

b: \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2-x}-\sqrt{4x^2+1}}{2x+3}\)

\(=\dfrac{\sqrt{x^2\left(1-\dfrac{1}{x}\right)}-\sqrt{x^2\left(4+\dfrac{1}{x^2}\right)}}{2x+3}\)

\(=\dfrac{-x\cdot\sqrt{1-\dfrac{1}{x}}+x\cdot\sqrt{4+\dfrac{1}{x^2}}}{x\left(2+\dfrac{3}{x}\right)}\)

\(=\dfrac{-\sqrt{1-\dfrac{1}{x}}+\sqrt{4+\dfrac{1}{x^2}}}{2+\dfrac{3}{x}}=\dfrac{-1+2}{2}=\dfrac{1}{2}\)

a: \(\lim\limits_{x\rightarrow+\infty}\left[x\left(\sqrt{x^2+2}-x\right)\right]\)
\(=\lim\limits_{x\rightarrow+\infty}\left[x\cdot\dfrac{x^2+2-x^2}{\sqrt{x^2+2}+x}\right]\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{2x}{\sqrt{x^2+2}+x}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{2}{\sqrt{1+\dfrac{2}{x^2}}+1}=\dfrac{2}{1+1}=\dfrac{2}{2}=1\)

b: \(\lim\limits_{x\rightarrow-\infty}\dfrac{3x^2-4x+6}{x-2}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2\left(3-\dfrac{4}{x}+\dfrac{6}{x^2}\right)}{x\left(1-\dfrac{2}{x}\right)}\)

\(=\lim\limits_{x\rightarrow-\infty}\left[x\cdot\dfrac{3-\dfrac{4}{x}+\dfrac{6}{x^2}}{1-\dfrac{2}{x}}\right]\)

\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow-\infty}x=-\infty\\\lim\limits_{x\rightarrow-\infty}\dfrac{3-\dfrac{4}{x}+\dfrac{6}{x^2}}{1-\dfrac{2}{x}}=\dfrac{3-0+0}{1-0}=\dfrac{3}{1}=3>0\end{matrix}\right.\)