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)

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

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

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

Lời giải:\(\lim\limits_{x\to 1}\left(\frac{m}{1-x^m}-\frac{n}{1-x^n}\right)=\lim\limits_{x\to 1}\left(\frac{m}{1-x^m}-\frac{1}{1-x}-\frac{n}{1-x^n}+\frac{1}{1-x}\right)\)

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

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

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

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

C2: Xài L'Hospital

\(\lim\limits_{x\rightarrow1}\dfrac{m-m.x^n-n+n.x^m}{1-x^m-x^n+x^{m+n}}=\lim\limits_{x\rightarrow1}\dfrac{m.n.x^{m-1}-m.n.x^{n-1}}{\left(m+n\right)x^{m+n-1}-m.x^{m-1}-n.x^{n-1}}\)

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

\(=\dfrac{m^2n-mn-mn^2+mn}{m^2+2mn-m+n^2-n-m^2+m-n^2+n}=\dfrac{mn\left(m-n\right)}{2mn}=\dfrac{m-n}{2}\)

1: \(A=\dfrac{x^2-\left(a+1\right)x+a}{x^3-a^3}\)

\(=\dfrac{x^2-xa-x+a}{\left(x-a\right)\left(x^2+ax+a^2\right)}\)

\(=\dfrac{\left(x-a\right)\left(x-1\right)}{\left(x-a\right)\left(x^2+ax+a^2\right)}=\dfrac{x-1}{x^2+ax+a^2}\)

\(lim_{x->a}A=lim_{x->a}\left(\dfrac{x-1}{x^2+ax+a^2}\right)\)

\(=\dfrac{a-1}{a^2+a^2+a^2}=\dfrac{a-1}{3a^2}\)

2: \(B=\dfrac{1}{1-x}-\dfrac{3}{1-x^3}\)

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

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

\(lim_{x->1}\left(B\right)=\dfrac{-1-2}{1^2+1+1}=\dfrac{-3}{3}=-1\)

3: \(C=\dfrac{\left(x+h\right)^3-x^3}{h}=\dfrac{\left(x+h-x\right)\left(x^2+2xh+h^2+x^2+hx+x^2\right)}{h}\)

\(=3x^2+3hx\)

\(lim_{h->0}\left(C\right)=3x^2+3\cdot0\cdot x=3x^2\)

Tui nghĩ cái này L'Hospital chứ giải thông thường là ko ổn :)

\(M=\lim\limits_{x\rightarrow0}\dfrac{\left(1+4x\right)^{\dfrac{1}{2}}-\left(1+6x\right)^{\dfrac{1}{3}}}{1-\cos3x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{2}\left(1+4x\right)^{-\dfrac{1}{2}}.4-\dfrac{1}{3}\left(1+6x\right)^{-\dfrac{2}{3}}.6}{3.\sin3x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-\dfrac{1}{4}.4\left(1+4x\right)^{-\dfrac{3}{2}}.4+\dfrac{2}{9}.6.6\left(1+6x\right)^{-\dfrac{5}{3}}}{3.3.\cos3x}\) 

Giờ thay x vô là được

\(N=\lim\limits_{x\rightarrow0}\dfrac{\left(1+ax\right)^{\dfrac{1}{m}}-\left(1+bx\right)^{\dfrac{1}{n}}}{\left(1+x\right)^{\dfrac{1}{2}}-1}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{m}.\left(1+ax\right)^{\dfrac{1}{m}-1}.a-\dfrac{1}{n}\left(1+bx\right)^{\dfrac{1}{n}-1}.b}{\dfrac{1}{2}\left(1+x\right)^{-\dfrac{1}{2}}}=\dfrac{\dfrac{a}{m}-\dfrac{b}{n}}{\dfrac{1}{2}}\)

\(V=\lim\limits_{x\rightarrow0}\dfrac{\left(1+mx\right)^n-\left(1+nx\right)^m}{\left(1+2x\right)^{\dfrac{1}{2}}-\left(1+3x\right)^{\dfrac{1}{3}}}=\lim\limits_{x\rightarrow0}\dfrac{n\left(1+mx\right)^{n-1}.m-m\left(1+nx\right)^{m-1}.n}{\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{1}{2}}.2-\dfrac{1}{3}\left(1+3x\right)^{-\dfrac{2}{3}}.3}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{n\left(n-1\right)\left(1+mx\right)^{n-2}.m-m\left(m-1\right)\left(1+nx\right)^{m-2}.n}{-\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{3}{2}}.2+\dfrac{2}{9}.3.3\left(1+3x\right)^{-\dfrac{5}{3}}}=....\left(thay-x-vo-la-duoc\right)\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x^2+3x+1\right)\left(\sqrt{3x+1}-2\right)+2\left(x^2+3x+1\right)-10}{x^2-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{3\left(x-1\right)\left(x^2+3x+1\right)}{\sqrt{3x+1}+2}+2\left(x-1\right)\left(x+4\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{3\left(x^2+3x+1\right)}{\sqrt{3x+1}+2}+2\left(x+4\right)}{x+1}=...\)

a/ \(=\lim\limits_{h\rightarrow0}\dfrac{2x^3+6x^2h+6xh^2+2h^3-2x^3}{h}\)

\(=\lim\limits_{h\rightarrow0}\dfrac{6xh^2+6x^2h+2h^3}{h}=\lim\limits_{h\rightarrow0}\left(6xh+6x^2+2h^2\right)=6x^2\)

b/ Xet day :\(S=x+x^2+....+x^{2021}\)

Day co \(\left\{{}\begin{matrix}u_1=x\\q=x\end{matrix}\right.\Rightarrow S=u_1.\dfrac{q^{2021}-1}{q-1}=x.\dfrac{x^{2021}-1}{x-1}\)

\(\Rightarrow\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^{2022}-x}{x-1}-2021}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{x^{2022}-x-2021x+2021}{\left(x-1\right)^2}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^{2022}}{x^2}-\dfrac{x}{x^2}-\dfrac{2021x}{x^2}+\dfrac{2021}{x^2}}{\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}+\dfrac{1}{x^2}}=\lim\limits_{x\rightarrow1}\dfrac{x^{2020}}{1}=1\)

Lam lai cau b, hinh nhu bi nham sang dang \(\dfrac{\infty}{\infty}\) roi

Xet day: \(S=x+x^2+...+x^{2021}\)

\(\Rightarrow S=x.\dfrac{x^{2021}-1}{x-1}=\dfrac{x^{2022}-x}{x-1}\)

\(\Rightarrow\lim\limits_{x\rightarrow1}\dfrac{x^{2022}-2022x+2021}{\left(x-1\right)^2}\)

L'Hospital: \(\Rightarrow...=\lim\limits_{x\rightarrow1}\dfrac{2022x^{2021}-2022}{2\left(x-1\right)}=\lim\limits_{x\rightarrow1}\dfrac{2022.2021.x^{2020}}{2}=2043231\)

Is that true :v?

Cau a co the xai L'Hospital cung ra:

L'Hospital: 

\(...=\lim\limits_{h\rightarrow0}\dfrac{6xh^2+6x^2h+2h^3}{h}=\lim\limits_{h\rightarrow0}\dfrac{6h^2+12xh+6x^2+12xh+6h^2}{1}=6x^2\)

Ta có \(\frac{x^n-nx+n-1}{\left(x-1\right)^2}=\frac{\left(x^n-1\right)-n\left(x-1\right)}{\left(x-1\right)^2}\)

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

Từ (1) suy ra :

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

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

         \(=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left[\left(x^{n-1}+x^{n-3}+.....+x+1\right)+.....+\left(x+1\right)+1\right]}{x-1}\)

         \(=\lim\limits_{x\rightarrow1}\left[1+\left(x+1\right)+....+\left(x^{n-2}+x^{n-3}+.....+x+1\right)\right]\)

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

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:

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