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

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

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

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

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:

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

Lời giải:

a. \(\lim\limits_{x\to 1+}(x^3+x+1)=3>0\)

\(\lim\limits_{x\to 1+}(x-1)=0\) và $x-1>0$ khi $x>1$

\(\Rightarrow \lim\limits_{x\to 1+}\frac{x^3+x+1}{x-1}=+\infty\)

b.

 \(\lim\limits_{x\to -1+}(3x+2)=-1<0\)

\(\lim\limits_{x\to -1+}(x+1)=0\) và $x+1>0$ khi $x>-1$

\(\Rightarrow \lim\limits_{x\to -1+}\frac{3x+2}{x+1}=-\infty\)

c.

\(\lim\limits_{x\to 2-}(x-15)=-17<0\)

\(\lim\limits_{x\to 2-}(x-2)=0\) và $x-2<0$ khi $x<2$

\(\Rightarrow \lim\limits_{x\to 2-}\frac{x-15}{x-2}=+\infty\)

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)

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

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

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

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

Lời giải:
\(L=\lim\limits_{x\to 1}\frac{\sqrt{2x-1}(\sqrt[3]{x+7}-2)+2(\sqrt{2x-1}-1)}{x(x-1)}=\lim\limits_{x\to 1}\frac{\sqrt{2x-1}.\frac{1}{\sqrt[3]{(x+7)^2}+2\sqrt[3]{x+7}+4}+4.\frac{1}{\sqrt{2x-1}+1}}{x}=\frac{25}{12}\)

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

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

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

\(=lim_{x\rightarrow1}\dfrac{-x^2+2x-1}{x\left(x-1\right)\left(\sqrt{2x-x^2}+1\right)}=lim_{x\rightarrow1}\dfrac{-\left(x-1\right)^2}{x\left(x-1\right)\left(\sqrt{2x-x^2}+1\right)}\)

\(=lim_{x\rightarrow1}-\dfrac{\left(x-1\right)}{x\left(\sqrt{2x-x^2}+1\right)}\)=0