a) lim= - 1/0 = - vô cùng

d) lim x(x^99-2)+1/ x(x^49-2)+1 =lim (x^99-2)/(x^49-2)=1

1/ \(\lim\limits_{x\rightarrow0^-}\left(\dfrac{x-2}{x^3}\right)=\lim\limits_{x\rightarrow0^-}\dfrac{2-x}{-x^3}=\dfrac{2}{0}=+\infty\)

2/ \(\lim\limits_{x\rightarrow1^+}\dfrac{\left(x^3-x^2\right)^{\dfrac{1}{2}}}{\left(x-1\right)^{\dfrac{1}{2}}+1-x}=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{1}{2}\left(x^3-x^2\right)^{-\dfrac{1}{2}}.\left(3x^2-2x\right)}{\dfrac{1}{2}\left(x-1\right)^{-\dfrac{1}{2}}-1}=0\)

3/ \(\lim\limits_{x\rightarrow1^+}\dfrac{1-\left(x^2+x+1\right)}{x^3-1}=\dfrac{1-3}{0}=-\infty\)

4/ \(\lim\limits_{x\rightarrow-\infty}\left(-\infty-\sqrt[3]{1+\infty}\right)=-\left(\infty+\infty\right)=-\infty?\) Cái này ko chắc :v

Có 2 cách làm

Cách 1: Phân tích bình thường

\(x+x^2+...+x^n-n=\left(x-1\right)+\left(x^2-1\right)+...+\left(x^n-1\right)\)

\(=\left(x-1\right)+\left(x-1\right)\left(x+1\right)+\left(x-1\right)\left(x^2+x+1\right)+....+\left(x-1\right)\left(x^{n-1}+x^{n-2}+...+1\right)\)

\(=\left(x-1\right)\left(1+x+1+x^2+x+1+...+x^{n-1}+x^{n-2}+...+1\right)\)

\(=\left(x-1\right)\left(1.n+\left(n-1\right)x+\left(n-2\right)x^2+...+\left(n-n+2\right)x^{n-2}+\left(n-n+1\right)x^{n-1}\right)\)

\(=\left(x-1\right)\left(n+\left(n-1\right)x+\left(n-2\right)x^2+...+2x^{n-2}+x^{n-1}\right)\)

\(\Rightarrow\lim\limits_{x\rightarrow1}\dfrac{x+x^2+...+x^n-n}{x-1}=\dfrac{\left(x-1\right)\left[n+\left(n-1\right)x+\left(n-2\right)x^2+...+2x^{n-2}+x^{n-1}\right]}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\left[n+\left(n-1\right)x+\left(n-2\right)x^2+...+2x^{n-2}+x^{n-1}\right]\)

\(=n+\left(n-1\right)+\left(n-2\right)+...+2+1=1+2+...+\left(n-1\right)+\left(n-2\right)=\dfrac{n\left(n+1\right)}{2}\)

Cách 2: Sử dụng L'Hospital

\(\lim\limits_{x\rightarrow1}\dfrac{x+x^2+...+x^n-n}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{1+2x+3x^2+...+nx^{n-1}}{1}=1+2.1+3.1+...+n=\dfrac{n\left(n+1\right)}{2}\)

\(a=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x^2-2x-2\right)}{\left(x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow1}\dfrac{x^2-2x-2}{x-3}=\dfrac{3}{2}\)

Câu b bạn coi lại đề, là \(x\rightarrow-1^-\) hay \(x\rightarrow1^-\) (đúng như đề thì ko phải dạng vô định, cứ thay số rồi bấm máy)

\(c=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)}{\left(x-3\right)\left(x-1\right)\left(\sqrt[3]{\left(x+5\right)^2}+2\sqrt[3]{x+5}+4\right)}\)

 \(=\lim\limits_{x\rightarrow3}\dfrac{1}{\left(x-1\right)\left(\sqrt[3]{\left(x+5\right)^2}+2\sqrt[3]{x+5}+4\right)}=\dfrac{1}{2.\left(4+4+4\right)}=...\)

a/ \(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}{\left(x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}{x-3}=....\)

Từ 2 câu kia lát tui làm, ăn cơm đã :D

Ta có: \(\dfrac{n}{1-x^n}-\dfrac{1}{1-x}=\dfrac{n-\left(1+x+x^2+...+x^{n-1}\right)}{1-x^n}\)

\(=\dfrac{1-x+1-x^2+...+1-x^{n-1}}{1-x^n}\)

\(=\dfrac{1+\left(1+x\right)+\left(1+x+x^2\right)+...+1+x+x^2+...+x^{n-2}}{1+x+x^2+...+x^{n-1}}\)

\(\Rightarrow\lim\limits_{x\rightarrow1}\left(\dfrac{n}{1-x^n}-\dfrac{1}{1-x}\right)=\dfrac{n-1}{2}\)

Lời giải:

\(\lim\limits_{x\to 1}\frac{x^n-nx+n-1}{(x-1)^2}=\lim\limits_{x\to 1}\frac{(x^n-1)-n(x-1)}{(x-1)^2}=\lim\limits_{x\to 1}\frac{(1+x+...+x^{n-1})-n}{x-1}\)

\(=\lim\limits_{x\to 1}\frac{(x-1)+(x^2-1)+...+(x^{n-1}-1)}{x-1}=\lim\limits_{x\to 1}[1+(x+1)+...+(1+x+...+x^{n-2})]\)

\(=\frac{n(n-1)}{2}\)

Bài này chắc chỉ xài L'Hopital chứ tách nhân tử thì không biết đến bao giờ mới xong:

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(n-1\right)x^{n-2}-\left(n+1\right)}{2\left(x-1\right)}=\dfrac{-2}{0}=-\infty\)

Chúng ta tính giới hạn sau:

\(\lim\limits_{x\rightarrow1}\dfrac{1-\sqrt[n]{x}}{1-x}\)

Cách đơn giản nhất là sử dụng L'Hopital:

\(\lim\limits_{x\rightarrow1}\dfrac{1-x^{\dfrac{1}{n}}}{1-x}=\lim\limits_{x\rightarrow1}\dfrac{-\dfrac{1}{n}x^{\dfrac{1}{n}-1}}{-1}=\dfrac{1}{n}\)

Phức tạp hơn thì tách mẫu theo hằng đẳng thức

\(=\lim\limits_{x\rightarrow1}\dfrac{1-\sqrt[x]{n}}{\left(1-\sqrt[n]{x}\right)\left(1+\sqrt[n]{x}+\sqrt[n]{x^2}+...+\sqrt[n]{x^{n-1}}\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{1+\sqrt[n]{x}+\sqrt[n]{x^2}+...+\sqrt[n]{x^{n-1}}}=\dfrac{1}{n}\)

Tóm lại ta có:

\(\lim\limits_{x\rightarrow1}\dfrac{1-\sqrt[n]{x}}{1-x}=\dfrac{1}{n}\)

Do đó:

\(I_1=\lim\limits_{x\rightarrow1}\left(\dfrac{1-\sqrt[2]{x}}{1-x}\right)\left(\dfrac{1-\sqrt[3]{x}}{1-x}\right)...\left(\dfrac{1-\sqrt[n]{x}}{1-x}\right)=\dfrac{1}{2}.\dfrac{1}{3}...\dfrac{1}{n}=\dfrac{1}{n!}\)

Câu 2 cũng vậy: L'Hopital hoặc tách hằng đẳng thức trâu bò (thôi L'Hopital đi cho đỡ sợ)

\(I_2=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+x^2}+x\right)^n-\left(\sqrt{1+x^2}-x\right)^n}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{n\left(\sqrt{1+x^2}+x\right)^{n-1}\left(\dfrac{x}{\sqrt{1+x^2}}+1\right)-n\left(\sqrt{1+x^2}-x\right)^{n-1}\left(\dfrac{x}{\sqrt{1+x^2}}-1\right)}{1}\)

\(=\dfrac{n.1-n\left(-1\right)}{1}=2n\)