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\(a=\lim\limits_{x\rightarrow a}\frac{\left(\sqrt{x}-\sqrt{a}\right)\left(x+\sqrt{ax}+a\right)}{\sqrt{x}-\sqrt{a}}=\lim\limits_{x\rightarrow a}\left(x+\sqrt{ax}+a\right)=3a\)
\(b=\lim\limits_{x\rightarrow1}\frac{x^{\frac{1}{n}}-1}{x^{\frac{1}{m}}-1}=\lim\limits_{x\rightarrow1}\frac{\frac{1}{n}x^{\frac{1-n}{n}}}{\frac{1}{m}x^{\frac{1-m}{m}}}=\frac{\frac{1}{n}}{\frac{1}{m}}=\frac{m}{n}\)
Ta có:
\(\lim\limits_{x\rightarrow1}\frac{1-\sqrt[n]{x}}{1-x}=\lim\limits_{x\rightarrow1}\frac{1-x^{\frac{1}{n}}}{1-x}=\lim\limits_{x\rightarrow1}\frac{-\frac{1}{n}x^{\frac{1-n}{n}}}{-1}=\frac{1}{n}\)
\(\Rightarrow c=\lim\limits_{x\rightarrow1}\frac{\left(1-\sqrt{x}\right)}{1-x}.\frac{\left(1-\sqrt[3]{x}\right)}{\left(1-x\right)}.\frac{\left(1-\sqrt[4]{x}\right)}{\left(1-x\right)}.\frac{\left(1-\sqrt[5]{x}\right)}{\left(1-x\right)}=\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}=\frac{1}{120}\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x\sqrt{x}}}}+1}=\frac{1}{2}\)
\(e=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+x}-1+1-\sqrt[3]{1+x}}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{x}{\sqrt{1+x}+1}+\frac{x}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\frac{1}{\sqrt{1+x}+1}+\frac{1}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}\right)=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
\(f=\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-3+3-\sqrt{x+7}}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2}\frac{\frac{8\left(x-2\right)}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{x-2}{3+\sqrt{x+7}}}{\left(x-1\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow2}\frac{\frac{8}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{1}{3+\sqrt{x+7}}}{x-1}=\frac{8}{27}-\frac{1}{6}=\frac{7}{54}\)
\(g=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{3x-2}-1+1-\sqrt{2x-1}}{\left(x-1\right)\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{3\left(x-1\right)}{\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1}-\frac{2\left(x-1\right)}{1+\sqrt{2x-1}}}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{\frac{3}{\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1}-\frac{2}{1+\sqrt{2x-1}}}{x^2+x+1}=0\)
\(h=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+9}+\sqrt[3]{2x-6}}{x^3+1}=\frac{\sqrt[3]{10}-\sqrt[3]{4}}{2}\)
\(a=\lim\limits_{x\rightarrow-\infty}\left(\frac{-x^2}{\sqrt[3]{\left(x^3-x^2\right)^2}+x\sqrt[3]{x^3-x^2}+x^2}\right)=\lim\limits_{x\rightarrow-\infty}\left(\frac{-1}{\sqrt[3]{\left(1-\frac{1}{x}\right)^3}+\sqrt[3]{1-\frac{1}{x}}+1}\right)=-\frac{1}{3}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{5x^2-8x}{\sqrt[3]{\left(x^3+5x^2\right)^2}+\sqrt[3]{\left(x^3+5x^2\right)\left(x^3+8x\right)}+\sqrt[3]{\left(x^3+8x\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\frac{5-\frac{8}{x}}{\sqrt[3]{\left(1+\frac{5}{x}\right)^2}+\sqrt[3]{\left(1+\frac{5}{x}\right)\left(1+\frac{8}{x^2}\right)}+\sqrt[3]{\left(1+\frac{8}{x^2}\right)^2}}=\frac{5}{3}\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{1}{\sqrt[3]{\left(x^3+1\right)^2}+x\sqrt[3]{x^3+1}+x^2}=\frac{1}{+\infty}=0\)
Bài 2:
\(a=\lim\limits_{x\rightarrow1^-}\left(\frac{1-x}{\left(x-1\right)\left(x+1\right)}\right)=\lim\limits_{x\rightarrow1^-}\frac{-1}{x+1}=-\frac{1}{2}\)
\(b=\lim\limits_{x\rightarrow1^+}\left(\frac{x^2+x+1-3}{\left(1-x\right)\left(x^2+x+1\right)}\right)=\lim\limits_{x\rightarrow1^+}\frac{\left(x-1\right)\left(x+2\right)}{\left(1-x\right)\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow1^+}\frac{-x-2}{x^2+x+1}=-1\)
\(c=\lim\limits_{x\rightarrow2^+}\left(\frac{1}{\left(x-1\right)\left(x-2\right)}-\frac{1}{\left(x-2\right)\left(x-3\right)}\right)=\lim\limits_{x\rightarrow2^+}\frac{-2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
Do \(x\rightarrow2^+\Rightarrow x>2\Rightarrow x-2>0\Rightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\rightarrow0^-\)
\(\Rightarrow\lim\limits_{x\rightarrow2^+}\frac{-2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=+\infty\)
Bài 2:
\(\lim\limits_{x\to 2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}=\lim\limits_{x\to 2}\frac{x^2-x-2}{(x+\sqrt{x+2}).\frac{4x+1-9}{\sqrt{4x+1}+3}}=\lim\limits_{x\to 2}\frac{(x-2)(x+1)(\sqrt{4x+1}+3)}{(x+\sqrt{x+2}).4(x-2)}=\lim\limits_{x\to 2}\frac{(x+1)(\sqrt{4x+1}+3)}{4(x+\sqrt{x+2})}=\frac{9}{8}\)
Bài 3:
\(\lim\limits_{x\to 0-}\frac{1-\sqrt[3]{x-1}}{x}=-\infty \)
\(\lim\limits_{x\to 0+}\frac{1-\sqrt[3]{x-1}}{x}=+\infty \)
Bài 4:
\(\lim\limits_{x\to -\infty}\frac{x^2-5x+1}{x^2-2}=\lim\limits_{x\to -\infty}\frac{1-\frac{5}{x}+\frac{1}{x^2}}{1-\frac{2}{x^2}}=1\)
Bài 5:
\(\lim\limits_{x\to +\infty}\frac{2x^2-4}{x^3+3x^2-9}=\lim\limits_{x\to +\infty}\frac{\frac{2}{x}-\frac{4}{x^3}}{1+\frac{3}{x}-\frac{9}{x^3}}=0\)
Bài 6:
\(\lim\limits_{x\to 2- }\frac{2x-1}{x-2}=\lim\limits_{x\to 2-}\frac{2(x-2)+3}{x-2}=\lim\limits_{x\to 2-}\left(2+\frac{3}{x-2}\right)=-\infty \)
Bài 7:
\(\lim\limits _{x\to 3+ }\frac{8+x-x^2}{x-3}=\lim\limits _{x\to 3+}\frac{1}{x-3}.\lim\limits _{x\to 3+}(8+x-x^2)=2(+\infty)=+\infty \)
Bài 8:
\(\lim\limits _{x\to -\infty}(8+4x-x^3)=\lim\limits _{x\to -\infty}(-x^3)=+\infty \)
Bài 9:
\(\lim\limits _{x\to -1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{x^2+3-4}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{(x-1)(x+1)}\)
\(\lim\limits _{x\to -1}\frac{\sqrt{x^2+3}+2}{(\sqrt[3]{x^2}-\sqrt[3]{x}+1)(x-1)}=\frac{-2}{3}\)
Bài 1:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)
Bài 2:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)
\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)
a) 
(x4 – x2 + x - 1) = x4(1 - 
) = +∞.
b) 
(-2x3 + 3x2 -5 ) = x3(-2 + 
) = +∞.
c) 
= 
= +∞.
d) \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+1}+x}{5-2x}=\lim\limits_{x\rightarrow+\infty}\dfrac{\left|x\right|\sqrt{1+\dfrac{1}{x^2}}+x}{5-2x}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{1+\dfrac{1}{x^2}}+x}{5-2x}\)\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x^2}}+1}{\dfrac{5}{x}-2}=-1\).
1.
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{x^2-3x-4}}{1-x^2}=\lim\limits_{x\to (-1)-}\frac{\sqrt{(x+1)(x-4)}}{(1-x)(1+x)}\)
\(=\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{(x-1)\sqrt{-(x+1)}}=-\infty\) do:
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{x-1}=\frac{-\sqrt{5}}{2}<0\) và \(\lim\limits_{x\to (-1)-}\frac{1}{\sqrt{-(x+1)}}=+\infty\)
2.
\(\lim\limits_{x\to 2+}\left(\frac{1}{x-2}-\frac{x+1}{\sqrt{x+2}-2}\right)=\lim\limits_{x\to 2+}\frac{1-(x+1)(\sqrt{x+2}+2)}{x-2}=-\infty\) do:
\(\lim\limits_{x\to 2+}\frac{1}{x-2}=+\infty\) và \(\lim\limits_{x\to 2+}[1-(x+1)(\sqrt{x+2}+2)]=-11<0\)
Bài 1:
\(\lim\limits _{x\to 1}\frac{4x^6-5x^5+x}{(1-x)^2}=\lim\limits _{x\to 1}\frac{x(x-1)^2(4x^3+3x^2+2x+1)}{(1-x)^2}\)
\(=\lim\limits _{x\to 1}x(4x^3+3x^2+2x+1)=1(4.1^3+3.1^2+2.1+1)=10\)
Bài 3:
\(\lim\limits _{x\to +\infty}[\sqrt{9x^2-4x+3}-(ax+b)]=0\)
\(\Rightarrow \lim\limits _{x\to +\infty}\frac{\sqrt{9x^2-4x+3}-(ax+b)}{x}=0\)
\(\Leftrightarrow \lim\limits _{x\to +\infty}\left(\sqrt{9-\frac{4}{x}+\frac{3}{x^2}}-a+\frac{b}{x}\right)=0\)
\(\Leftrightarrow a=3\)
Thay $a=3$ vào đk ban đầu:
\(\lim\limits _{x\to +\infty}[\sqrt{9x^2-4x+3}-3x-b]=0\)
\(\Leftrightarrow \lim\limits _{x\to +\infty} (\sqrt{9x^2-4x+3}-3x)=b\)
\(\Leftrightarrow \lim\limits _{x\to +\infty}\frac{-4x+3}{\sqrt{9x^2-4x+3}+3x}=b\)
\(\Leftrightarrow \lim\limits _{x\to +\infty}\frac{-4+\frac{3}{x}}{\sqrt{9-\frac{4}{x}+\frac{3}{x}}+3}=b\)
\(\Leftrightarrow \frac{-4}{6}=b\Leftrightarrow b=-\frac{2}{3}\)
Mấy câu này bạn cần giải theo kiểu trắc nghiệm hay tự luận nhỉ?
Em cần kiểu tự luận ạ
Làm tự luận thì hơi tốn thời gian đấy (đi thi sẽ không bao giờ đủ thời gian đâu)
Câu 1:
Kiểm tra lại đề, \(\lim\limits_{x\rightarrow1}\dfrac{1}{\left(\sqrt[]{x}-1\right)g\left(x\right)}\) hay một trong 2 giới hạn sau: \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[]{x}-1}{g\left(x\right)}\) hoặc \(\lim\limits_{x\rightarrow1}\dfrac{g\left(x\right)}{\sqrt[]{x}-1}\)
Vì đúng như đề của bạn thì \(\lim\limits_{x\rightarrow1}\dfrac{1}{\left(\sqrt[]{x}-1\right)g\left(x\right)}=\dfrac{1}{0}=\infty\), cả \(g\left(x\right)\) lẫn \(\sqrt{x}-1\) đều tiến tới 0 khi x dần tới 1
Bài 2:
\(x^3-3x+2=\left(x-1\right)^2\left(x+2\right)\) có nghiệm kép \(x=1\)
Do đó giới hạn \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[]{2ax^2+30}-bx-5}{x^3-3x+2}\) hữu hạn khi và chỉ khi \(\sqrt{2ax^2+30}-bx-5=0\) (1) cũng có ít nhất nghiệm kép \(x=1\)
Thay \(x=1\) vào (1) ta được:
\(\sqrt{2a+30}=b+5\Rightarrow2a+30=\left(b+5\right)^2\) với \(b\ge-5\)
\(\Leftrightarrow2a=b^2+10b-5\)
Tiếp tục thế lên (1):
\(\Rightarrow\sqrt{\left(b^2+10b-5\right)x^2+30}=bx+5\)
\(\Rightarrow\left(b^2+10b-5\right)x^2+30=b^2x^2+10bx+25\)
\(\Rightarrow\left(2b-1\right)x^2-2bx+1=0\)
\(\Rightarrow2bx\left(x-1\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2bx-x-1\right)=0\) (2)
Để (2) có nghiệm kép \(x=1\Rightarrow2bx-x-1=0\) phải có nghiệm \(x=1\)
\(\Rightarrow2b-2=0\Rightarrow b=1\Rightarrow a=\dfrac{b^2+10b-5}{2}=3\)
Khi đó:
\(c=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[]{6x^2+30}-x-5}{x^3-3x+2}=\lim\limits_{x\rightarrow1}\dfrac{5\left(x-1\right)^2}{\left(x-1\right)^2\left(x+2\right)\left(\sqrt[]{6x^2+30}+x+5\right)}=\dfrac{5}{36}\)
3.
\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{4x^2+ax}+2x+\sqrt[3]{8x^3+2bx^2+3}-2x\right)\)
\(=\lim\limits_{x\rightarrow-\infty}\left(\dfrac{ax}{\sqrt{4x^2+ax}-2x}+\dfrac{2bx^2+3}{\sqrt[3]{\left(8x^3+2bx^2+3\right)^2}+2x\sqrt[3]{8x^3+2bx^2+3}+4x^2}\right)\)
\(=\lim\limits_{x\rightarrow-\infty}\left(\dfrac{a}{-\sqrt{4+\dfrac{a}{x}}-2}+\dfrac{2b+\dfrac{3}{x^2}}{\sqrt[3]{\left(8+\dfrac{2b}{x}+\dfrac{3}{x^3}\right)^2}+2\sqrt[3]{8+\dfrac{2b}{x}+\dfrac{3}{x^3}}+4}\right)\)
\(=-\dfrac{a}{4}+\dfrac{b}{6}=\dfrac{7}{3}\)
\(\Leftrightarrow-3a+2b=28\)
Đề bài không chính xác, tới đây có vô số bộ nguyên dương (a;b) thỏa mãn, ví dụ \(\left(a;b\right)=\left(2;17\right);\left(4;20\right);\left(6;23\right)\) vân vân đều thỏa mãn giả thiết (bạn có thể kiểm tra lại điều này dễ dàng bằng casio)
Do đó có vô số giá trị P
4.
Nếu \(c\le0\Rightarrow\) giới hạn đã cho trở thành:
\(\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{a+\dfrac{b}{x}}-c\right)=+\infty.\left(a-c\right)=+\infty\) (ko thỏa mãn là giá trị hữu hạn)
\(\Rightarrow c>0\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{ax^2+bx}-cx\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(a-c^2\right)x^2+bx}{\sqrt{ax^2+bx}+cx}=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(a-c^2\right)x+b}{\sqrt{a+\dfrac{b}{x}}+c}\) (1)
Nếu \(a-c^2\ne0\Rightarrow\) giới hạn đã cho bằng vô cực (ktm)
\(\Rightarrow a-c^2=0\) , kết hợp \(a+c^2=2\Rightarrow a=c=1\)
Thế vào (1):
\(\lim\limits_{x\rightarrow+\infty}\dfrac{b}{\sqrt{1}+1}=-3\Rightarrow b=-6\)
47.
Hàm số liên tục tại \(x=\dfrac{1}{2}\) khi và chỉ khi:
\(\lim\limits_{x\rightarrow\dfrac{1}{2}}f\left(x\right)=f\left(\dfrac{1}{2}\right)\Leftrightarrow\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{\sqrt{ax^2+1}-bx-2}{4x^3-3x+1}=\dfrac{c}{2}\) hữu hạn
Mà \(4x^3-3x+1=\left(x+1\right)\left(2x-1\right)^2\) có nghiệm kép \(x=\dfrac{1}{2}\)
\(\Rightarrow\sqrt{ax^2+1}-bx-2\) cũng có ít nhất nghiệm kép \(x=\dfrac{1}{2}\)
Lặp lại quy trình tính như câu 2:
Thay \(x=\dfrac{1}{2}\Rightarrow\sqrt{\dfrac{a}{4}+1}=\dfrac{b}{2}+2\Rightarrow a=b^2+8b+12\)
\(\sqrt{\left(b^2+8b+12\right)x^2+1}=bx+2\) có nghiệm kép \(x=\dfrac{1}{2}\)
\(\Rightarrow\left(b^2+8b+12\right)x^2+1=b^2x^2+4bx+4\)
\(\Rightarrow8bx^2-4bx+12x^2-3=0\)
\(\Rightarrow\left(2x-1\right)\left(4bx+6x+3\right)=0\)
\(\Rightarrow4bx+6x+3=0\) có nghiệm \(x=\dfrac{1}{2}\Rightarrow b=-3\)
\(\Rightarrow a=-3\)
\(c=2.\left(\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{\sqrt{-3x^2+1}+3x-2}{4x^3-3x+1}\right)=-4\)
49.
Câu này chúng ta không nên liên hợp 1 chút xíu nào, hãy sử dụng L'Hopital:
\(\lim\limits_{x\rightarrow1}\dfrac{x^{2020}+x-2}{\sqrt{2021x+1}-\sqrt{x+2021}}=\lim\limits_{x\rightarrow1}\dfrac{2020x^{2019}+1}{\dfrac{2021}{2\sqrt{2021x+1}}-\dfrac{1}{2\sqrt{x+2021}}}\)
\(=\dfrac{2021}{\dfrac{2021}{2\sqrt{2022}}-\dfrac{1}{2\sqrt{2022}}}=\dfrac{2021\sqrt{2022}}{1010}\)
\(\Rightarrow\left\{{}\begin{matrix}m=2021\\n=2022\end{matrix}\right.\)
Vậy nếu câu 1 em sửa lại đề bài thành tính \(\lim\limits_{x\rightarrow1}\dfrac{g\left(x\right)}{\sqrt{x}-1}\) thì làm như thế nào ạ