Tính: \(I=\lim\limits_{x\rightarrow1}\left(\dfrac{n}{1-x^n}-\dfrac{m}{1-x^m}\right)\) với m,n là các số nguyên.
Tính giới hạn :
\(\lim\limits_{x\rightarrow1}\left(\dfrac{m}{1-x^m}-\dfrac{n}{1-x^n}\right)\)
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
\(\lim\limits_{x\rightarrow1}\dfrac{m-m.x^n-n+n.x^m}{1-x^m-x^n+x^{m+n}}=\lim\limits_{x\rightarrow1}\dfrac{m.n.x^{m-1}-m.n.x^{n-1}}{\left(m+n\right)x^{m+n-1}-m.x^{m-1}-n.x^{n-1}}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{m.n.\left(m-1\right).x^{m-2}-m.n.\left(n-1\right).x^{n-2}}{\left(m+n\right).\left(m+n-1\right)x^{m+n-2}-m\left(m-1\right)x^{m-2}-n\left(n-1\right)x^{n-2}}\)
\(=\dfrac{m^2n-mn-mn^2+mn}{m^2+2mn-m+n^2-n-m^2+m-n^2+n}=\dfrac{mn\left(m-n\right)}{2mn}=\dfrac{m-n}{2}\)
Tính các giới hạn sau:\(I_1=\lim\limits_{x\rightarrow1}\dfrac{\left(1-\sqrt{x}\right)\left(1-\sqrt[3]{x}\right)....\left(1-\sqrt[n]{x}\right)}{\left(1-x\right)^{n-1}}\)
\(I_2=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+x^2}+x\right)^n-\left(\sqrt{1+x^2}-x\right)^n}{x}\)
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)
\(I_2=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+x^2}+x\right)^n-\left(\sqrt{1+x^2}-x\right)^n}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{n\left(\sqrt{1+x^2}+x\right)^{n-1}\left(\dfrac{x}{\sqrt{1+x^2}}+1\right)-n\left(\sqrt{1+x^2}-x\right)^{n-1}\left(\dfrac{x}{\sqrt{1+x^2}}-1\right)}{1}\)
\(=\dfrac{n.1-n\left(-1\right)}{1}=2n\)
Tính \(\lim\limits_{x\rightarrow1}\left(\dfrac{n}{1-x^n}-\dfrac{1}{1-x}\right)\)
Ta có: \(\dfrac{n}{1-x^n}-\dfrac{1}{1-x}=\dfrac{n-\left(1+x+x^2+...+x^{n-1}\right)}{1-x^n}\)
\(=\dfrac{1-x+1-x^2+...+1-x^{n-1}}{1-x^n}\)
\(=\dfrac{1+\left(1+x\right)+\left(1+x+x^2\right)+...+1+x+x^2+...+x^{n-2}}{1+x+x^2+...+x^{n-1}}\)
\(\Rightarrow\lim\limits_{x\rightarrow1}\left(\dfrac{n}{1-x^n}-\dfrac{1}{1-x}\right)=\dfrac{n-1}{2}\)
Tính giới hạn :
\(\lim\limits_{x\rightarrow1}\dfrac{x^{n-1}-\left(n+1\right)x+n}{\left(x-1\right)^2}\)
Bài này chắc chỉ xài L'Hopital chứ tách nhân tử thì không biết đến bao giờ mới xong:
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(n-1\right)x^{n-2}-\left(n+1\right)}{2\left(x-1\right)}=\dfrac{-2}{0}=-\infty\)
Tính các giới hạn sau:\(M=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{1-cos3x}\)
\(N=\lim\limits_{X\rightarrow0}\dfrac{\sqrt[m]{1+ax}-\sqrt[n]{1+bx}}{\sqrt{1+x}-1}\)
\(V=\lim\limits_{x\rightarrow0}\dfrac{\left(1+mx\right)^n-\left(1+nx\right)^m}{\sqrt{1+2x}-\sqrt[3]{1+3x}}\)
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
\(N=\lim\limits_{x\rightarrow0}\dfrac{\left(1+ax\right)^{\dfrac{1}{m}}-\left(1+bx\right)^{\dfrac{1}{n}}}{\left(1+x\right)^{\dfrac{1}{2}}-1}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{m}.\left(1+ax\right)^{\dfrac{1}{m}-1}.a-\dfrac{1}{n}\left(1+bx\right)^{\dfrac{1}{n}-1}.b}{\dfrac{1}{2}\left(1+x\right)^{-\dfrac{1}{2}}}=\dfrac{\dfrac{a}{m}-\dfrac{b}{n}}{\dfrac{1}{2}}\)
\(V=\lim\limits_{x\rightarrow0}\dfrac{\left(1+mx\right)^n-\left(1+nx\right)^m}{\left(1+2x\right)^{\dfrac{1}{2}}-\left(1+3x\right)^{\dfrac{1}{3}}}=\lim\limits_{x\rightarrow0}\dfrac{n\left(1+mx\right)^{n-1}.m-m\left(1+nx\right)^{m-1}.n}{\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{1}{2}}.2-\dfrac{1}{3}\left(1+3x\right)^{-\dfrac{2}{3}}.3}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{n\left(n-1\right)\left(1+mx\right)^{n-2}.m-m\left(m-1\right)\left(1+nx\right)^{m-2}.n}{-\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{3}{2}}.2+\dfrac{2}{9}.3.3\left(1+3x\right)^{-\dfrac{5}{3}}}=....\left(thay-x-vo-la-duoc\right)\)
\(\lim\limits_{x\rightarrow1}\left(\dfrac{n}{1-x^n}-\dfrac{1}{1-x}\right)\)
Tính các giới hạn
a) \(\lim\limits_{x\rightarrow\infty}\dfrac{a_0x^m+a_1x^{m-1}+a_2x^{m-2}+...+a_m}{b_0x^n+b_1x^{n-1}+b_2x^{n-2}+...+b_n}\)
b) \(\lim\limits_{x\rightarrow+\infty}\dfrac{\left(x-\sqrt{x^2-1}\right)^n+\left(x+\sqrt{x^2-1}\right)^n}{x^n}\)
Tính giới hạn :
\(\lim\limits_{x\rightarrow1}\dfrac{x^n-nx+n-1}{\left(x-1\right)^2}\)
Lời giải:
\(\lim\limits_{x\to 1}\frac{x^n-nx+n-1}{(x-1)^2}=\lim\limits_{x\to 1}\frac{(x^n-1)-n(x-1)}{(x-1)^2}=\lim\limits_{x\to 1}\frac{(1+x+...+x^{n-1})-n}{x-1}\)
\(=\lim\limits_{x\to 1}\frac{(x-1)+(x^2-1)+...+(x^{n-1}-1)}{x-1}=\lim\limits_{x\to 1}[1+(x+1)+...+(1+x+...+x^{n-2})]\)
\(=\frac{n(n-1)}{2}\)
cho m, n là các số thực khác 0. nếu \(\lim\limits_{x\rightarrow1}\dfrac{x^2+mx+n}{x-1}=3\) thì m.n=?
\(\lim\limits_{x\rightarrow1}\dfrac{x^2+mx+n}{x-1}\) hữu hạn khi \(x^2+mx+n=0\) có nghiệm \(x=1\)
\(\Rightarrow1+m+n=0\Rightarrow n=-m-1\)
\(\lim\limits_{x\rightarrow1}\dfrac{x^2+mx-m-1}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+m+1\right)}{x-1}=\lim\limits_{x\rightarrow1}\left(x+m+1\right)=m+2\)
\(\Rightarrow m+2=3\Rightarrow m=1\Rightarrow n=-2\)
\(\Rightarrow mn=-2\)