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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 \(x=1\)
\(\Rightarrow1+m+n=0\Rightarrow n=-m-1\)
\(\lim\limits_{x\rightarrow1}\dfrac{x^2+mx-m-1}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+m+1\right)}{x-1}=\lim\limits_{x\rightarrow1}\left(x+m+1\right)=m+2\)
\(\Rightarrow m+2=3\Rightarrow m=1\Rightarrow n=-2\)
\(\Rightarrow mn=-2\)
a) \(lim\dfrac{-2n+1}{n}=lim\dfrac{\dfrac{-2n}{n}+\dfrac{1}{n}}{\dfrac{n}{n}}=lim\dfrac{-2+\dfrac{1}{n}}{1}=\dfrac{lim\left(-2\right)+\dfrac{lim1}{n}}{lim1}=\dfrac{-2+0}{1}=-\dfrac{2}{1}=-2\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-\left(x+8\right)}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{3+\sqrt{x+8}}=\dfrac{1}{3+\sqrt{1+8}}=\dfrac{1}{3+3}=\dfrac{1}{9}\)
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:
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 $x<2$
\(\Rightarrow \lim\limits_{x\to 2-}\frac{x-15}{x-2}=+\infty\)
\(\lim\limits_{x\rightarrow1}\dfrac{\left(x^2-3x+2\right)\left(x+\sqrt{5x-4}\right)}{\left(x^2-5x+4\right)\left(x+2+\sqrt{7x+2}\right)}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x-2\right)\left(x+\sqrt{5x-4}\right)}{\left(x-1\right)\left(x-5\right)\left(x+2+\sqrt{7x+2}\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-2\right)\left(x+\sqrt{5x-4}\right)}{\left(x-5\right)\left(x+2+\sqrt{7x+2}\right)}=\dfrac{1}{12}\)
\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=12\end{matrix}\right.\)
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)
\(I_2=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+x^2}+x\right)^n-\left(\sqrt{1+x^2}-x\right)^n}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{n\left(\sqrt{1+x^2}+x\right)^{n-1}\left(\dfrac{x}{\sqrt{1+x^2}}+1\right)-n\left(\sqrt{1+x^2}-x\right)^{n-1}\left(\dfrac{x}{\sqrt{1+x^2}}-1\right)}{1}\)
\(=\dfrac{n.1-n\left(-1\right)}{1}=2n\)
Lại xài L'Hopital:
\(=\lim\limits_{x\rightarrow1}\dfrac{1+2x+3x^2+...+nx^{n-1}}{1+2x+3x^2+...+mx^{m-1}}=\dfrac{1+2+...+n}{1+2+...+m}=\dfrac{n\left(n+1\right)}{m\left(m+1\right)}\)
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
\(N=\lim\limits_{x\rightarrow0}\dfrac{\left(1+ax\right)^{\dfrac{1}{m}}-\left(1+bx\right)^{\dfrac{1}{n}}}{\left(1+x\right)^{\dfrac{1}{2}}-1}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{m}.\left(1+ax\right)^{\dfrac{1}{m}-1}.a-\dfrac{1}{n}\left(1+bx\right)^{\dfrac{1}{n}-1}.b}{\dfrac{1}{2}\left(1+x\right)^{-\dfrac{1}{2}}}=\dfrac{\dfrac{a}{m}-\dfrac{b}{n}}{\dfrac{1}{2}}\)
\(V=\lim\limits_{x\rightarrow0}\dfrac{\left(1+mx\right)^n-\left(1+nx\right)^m}{\left(1+2x\right)^{\dfrac{1}{2}}-\left(1+3x\right)^{\dfrac{1}{3}}}=\lim\limits_{x\rightarrow0}\dfrac{n\left(1+mx\right)^{n-1}.m-m\left(1+nx\right)^{m-1}.n}{\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{1}{2}}.2-\dfrac{1}{3}\left(1+3x\right)^{-\dfrac{2}{3}}.3}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{n\left(n-1\right)\left(1+mx\right)^{n-2}.m-m\left(m-1\right)\left(1+nx\right)^{m-2}.n}{-\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{3}{2}}.2+\dfrac{2}{9}.3.3\left(1+3x\right)^{-\dfrac{5}{3}}}=....\left(thay-x-vo-la-duoc\right)\)
Cách 1 là quy đồng sau đó L'Hopital khoảng 2-3 lần gì đó là hết dạng vô định (đoán thế vì dạng vô định đa thức này nếu quy đồng sẽ luôn dùng L'Hopital giết được, vấn đề chỉ là L'Hopital bao nhiêu lần)
Cách 2:
Đặt \(y=\dfrac{1}{x}\), khi đó:
\(I=\lim\limits_{y\rightarrow1}\left(\dfrac{n}{1-\dfrac{1}{y^n}}-\dfrac{m}{1-\dfrac{1}{y^m}}\right)=\lim\limits_{y\rightarrow1}\left(\dfrac{n.y^n}{y^n-1}-\dfrac{m.y^m}{y^m-1}\right)\)
\(=\lim\limits_{x\rightarrow1}\left(\dfrac{n.x^n}{x^n-1}-\dfrac{m.x^m}{x^m-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{n\left(x^n-1+1\right)}{x^n-1}-\dfrac{m\left(x^m-1+1\right)}{x^m-1}\right)\)
\(=\lim\limits_{x\rightarrow1}\left(n+\dfrac{n}{x^n-1}-m-\dfrac{m}{x^m-1}\right)\)
\(=n-m-\lim\limits_{x\rightarrow1}\left(\dfrac{n}{1-x^n}-\dfrac{m}{1-x^m}\right)=n-m-I\)
Hay \(I=n-m-I\Rightarrow2I=n-m\)
\(\Rightarrow I=\dfrac{n-m}{2}\)
Do giới hạn hữu hạn nên \(x^2+mx+n=0\) có nghiệm \(x=1\)
\(\Rightarrow1+m+n=0\Rightarrow n=-m-1\)
\(\lim\limits_{x\rightarrow1}\dfrac{x^2+mx-m-1}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)+m\left(x-1\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1+m\right)}{x-1}=\lim\limits_{x\rightarrow1}\left(x+1+m\right)=m+2\)
\(\Rightarrow m+2=3\Rightarrow m=1\Rightarrow n=-2\)