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\(1=\lim\limits_{x\rightarrow0}\frac{\sqrt{x+4}-2}{2x}=\lim\limits_{x\rightarrow0}\frac{x}{2x}.\frac{1}{\sqrt{x+4}+2}=\lim\limits_{x\rightarrow0}\frac{1}{2\left(\sqrt{x+4}+2\right)}=\frac{1}{2\left(\sqrt{4}+2\right)}\)
\(2=\lim\limits_{x\rightarrow1}\frac{\sqrt{x+3}-2}{x-1}=\lim\limits_{x\rightarrow1}\frac{x-1}{x-1}.\frac{1}{\sqrt{x+3}+2}=\lim\limits_{x\rightarrow1}\frac{1}{\sqrt{x+3}+2}=\frac{1}{\sqrt{1+3}+2}\)
\(3=\lim\limits_{x\rightarrow3}\frac{\sqrt{2x+3}-x}{\left(x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow3}\frac{2x+3-x^2}{\left(x-1\right)\left(x-3\right)}.\frac{1}{\sqrt{2x+3}+x}\)
\(=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(3-x\right)}{\left(x-1\right)\left(x-3\right)}.\frac{1}{\sqrt{2x+3}+x}=\lim\limits_{x\rightarrow3}\frac{x+1}{\left(1-x\right)\left(\sqrt{2x+3}+x\right)}=\frac{3+1}{\left(1-3\right)\left(\sqrt{9}+3\right)}\)
\(4=\lim\limits_{x\rightarrow2}\frac{\left(x-2\right)\left(2x-1\right)}{\left(x+1\right)^2\left(x-2\right)}=\lim\limits_{x\rightarrow2}\frac{2x-1}{\left(x+1\right)^2}=\frac{4-1}{\left(2+1\right)^2}\)
P/s: lần sau bạn sử dụng tính năng gõ công thức ở kí hiệu \(\sum\) góc trên cùng bên trái khung soạn thảo ấy, khó nhìn đề quá chẳng muốn làm
1/ \(=\lim\limits_{x\rightarrow0}\dfrac{3\left(1+3x\right)^2.3+4.4\left(1-4x\right)^3}{1}=...\left(thay-x-vo\right)\)
2/ \(=\lim\limits_{x\rightarrow2}\dfrac{2.2.x-5}{3x^2-3}=\dfrac{4.2-5}{3.4-3}=\dfrac{1}{3}\)
3/ \(=\lim\limits_{x\rightarrow1}\dfrac{4x^3-3}{3x^2+2}=\dfrac{4.1-3}{3.1-2}=1\)
Xai L'Hospital nhe :v
câu 1 bạn lm kiểu j vậy chả hiểu luôn bạn có thể lm lại chi tiết hơn dc ko
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:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)
\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)
Do quá làm biếng dùng Hoocne tách nhân tử nên chúng ta sẽ sử dụng L'Hopital:
\(\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{x^2-2x+1}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2x-2}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=\frac{120-100}{2}=10\)
\(\lim\limits_{x\rightarrow-3}\frac{x^4-6x^2-27}{x^3+3x^2+x+3}=\lim\limits_{x\rightarrow-3}\frac{4x^3-12x}{3x^2+6x+1}=\frac{-36}{5}\)
\(\lim\limits_{x\rightarrow-2}\frac{2x^3+x^2+12}{-x^2-6x-8}=\lim\limits_{x\rightarrow-2}\frac{6x^2+2x}{-2x-6}=-10\)
\(\lim\limits_{x\rightarrow-2}\frac{-2x^3+x-14}{-2x^3-x^2-12}=\lim\limits_{x\rightarrow-2}\frac{-6x^2+1}{-6x^2-2x}=\frac{23}{20}\)
Con cuối ko phải tích phân dạng vô định \(\frac{0}{0}\) bạn cứ thế thẳng -2 vào là được
\(A=\lim\limits_{x\rightarrow2}\frac{\left(x-2\right)\left(2x-1\right)}{x-2}=\lim\limits_{x\rightarrow2}\left(2x-1\right)=3\)
\(B=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x^2-2x+3\right)}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\frac{x^2-2x+3}{x+1}=\frac{1-2+3}{1+1}=1\)
\(C=\lim\limits_{x\rightarrow2}\frac{x^2+2x}{x^2+4x+4}=\frac{4+4}{4+8+4}=\frac{1}{2}\)
\(D=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x^2-1\right)}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow1}\frac{x^2-1}{x-2}=\frac{0}{-1}=0\)
\(E=\lim\limits_{x\rightarrow1}\frac{x^3-5x^2+3x+9}{x^4-8x^4-9}=\frac{1-5+3+9}{1-8-9}=-\frac{1}{2}\)
\(F=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x-1\right)\left(x^2+1\right)}{\left(x+1\right)\left(x^2-3x+3\right)}=\lim\limits_{x\rightarrow-1}\frac{\left(x-1\right)\left(x^2+1\right)}{x^2-3x+3}=\frac{-2.2}{1+3+3}=-\frac{2}{5}\)
\(G=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(2x+1\right)}=\lim\limits_{x\rightarrow1}\frac{x+3}{2x+1}=\frac{4}{3}\)
\(H=\lim\limits_{x\rightarrow-2}\frac{\left(x+2\right)\left(x-1\right)^2}{\left(2-x\right)\left(x+2\right)}=\lim\limits_{x\rightarrow-2}\frac{\left(x-1\right)^2}{2-x}=\frac{9}{4}\)
\(I=\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+1}{x^2-1}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4}{2x}=\frac{24-25}{2}=-\frac{1}{2}\)
\(K=\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)
\(a=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-1+1-\sqrt[3]{2x+1}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x}{\sqrt[]{4x+1}+1}+\dfrac{-2x}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{4x+1}+1}+\dfrac{-2}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}\right)=...\)
\(b=\lim\limits_{x\rightarrow1}\dfrac{4\left(x-1\right)\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(x-1\right)\left(\sqrt[]{4x+5}+3\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{4\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(\sqrt[]{4x+5}+3\right)}=...\)
\(c=\lim\limits_{x\rightarrow-1}\dfrac{\left(2x+3\right)^{\dfrac{1}{4}}+\left(2+3x\right)^{\dfrac{1}{3}}}{\left(x+2\right)^{\dfrac{1}{2}}-1}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{1}{2}\left(2x+3\right)^{-\dfrac{3}{4}}+\left(2+3x\right)^{-\dfrac{2}{3}}}{\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}=3\)
a.
\(\lim\limits_{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}=\lim_{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{(x-1)^3(3x+1)}=\lim\limits _{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{3x+1}.\lim\limits_{x\to 1+}\frac{1}{(x-1)^3}\)
\(=\frac{1}{4}.(+\infty)=+\infty \)
Hoàn toàn tương tự:
\(\lim\limits_{x\to 1-}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}=-\infty \)
Do đó: \(\lim\limits_{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\neq \lim\limits_{x\to 1-}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\) nên không tồn tại \(\lim\limits_{x\to 1}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\)
b.
\(\lim\limits_{x\to 1+}\frac{x^3-3x^2+2}{x^4-4x+3}=\lim\limits_{x\to 1+}\frac{(x-1)(x^2-2x-2)}{(x-1)^2(x^2+2x+3)}=\lim\limits_{x\to 1+}\frac{x^2-2x-2}{(x-1)(x^2+2x+3)}\)
\(=\lim\limits_{x\to 1+}\frac{x^2-2x-2}{x^2+2x+3}.\lim\limits_{x\to 1+}\frac{1}{x-1}=\frac{-1}{2}.(+\infty)=-\infty \)
Tương tự \(\lim\limits_{x\to 1-}\frac{x^3-3x^2+2}{x^4-4x+3}=+\infty \)
Do đó không tồn tại \(\lim\limits_{x\to 1}\frac{x^3-3x^2+2}{x^4-4x+3}\)
c.
\(\lim\limits_{x\to 1}\frac{x^3-2x-1}{x^5-2x-1}=\frac{1^3-2.1-1}{1^5-2.1-1}=1\)
d.
\(\lim\limits_{x\to -1}\frac{(x+2)^2-1}{x^2-1}=\lim\limits_{x\to -1}\frac{(x+2-1)(x+2+1)}{(x-1)(x+1)}=\lim\limits_{x\to -1}\frac{x+3}{x-1}=-1\)
1/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}.\sqrt[4]{1+8x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}-1}{x}\)
Liên hợp dài quá ko muốn gõ tiếp, bạn tự đặt nhân tử chung rồi liên hợp nhé, kết quả ra 5
2/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{1+7x}-2-\left(x^3-3x+2\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{7\left(x-1\right)}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)^2\left(x+2\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{7}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)\left(x+2\right)=\dfrac{7}{12}\)
3/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x^3-x^2+1}{2x^2+3x-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{x-1+\dfrac{1}{x^2}}{2+\dfrac{3}{x}-\dfrac{1}{x^2}}=-\infty\)
4/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}}{\sqrt{4x+1}}=\lim\limits_{x\rightarrow+\infty}\dfrac{1+\dfrac{1}{\sqrt[6]{x}}+\dfrac{1}{\sqrt[4]{x}}}{\sqrt{4+\dfrac{1}{x}}}=\dfrac{1}{\sqrt{4}}=\dfrac{1}{2}\)
5/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+\sqrt{x^2+2}}{\sqrt[3]{8x^3+x^2+1}}=\lim\limits_{x\rightarrow-\infty}\dfrac{1-\sqrt{1+\dfrac{2}{x^2}}}{\sqrt[3]{8+\dfrac{1}{x}+\dfrac{1}{x^3}}}=\dfrac{1-1}{\sqrt[3]{8}}=0\)
6/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+3x-7}}{\sqrt[3]{27x^3+5x^2+x-4}}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4+\dfrac{3}{x}-\dfrac{7}{x^2}}}{\sqrt[3]{27+\dfrac{5}{x}+\dfrac{1}{x^2}-\dfrac{4}{x^3}}}=\dfrac{-\sqrt{4}}{\sqrt[3]{27}}=\dfrac{-2}{3}\)
a. \(lim_{x\rightarrow3}\dfrac{x^3-27}{3x^2-5x-2}=\dfrac{3^3-27}{3.3^2-5.3-2}=\dfrac{0}{10}=0\)
b. \(lim_{x\rightarrow2}\dfrac{\sqrt{x+2}-2}{4x^2-3x-2}=\dfrac{\sqrt{2+2}-2}{4.2^2-3.2-2}=\dfrac{0}{8}=0\)
c. \(lim_{x\rightarrow1}\dfrac{1-x^2}{x^2-5x+4}=lim_{x\rightarrow1}\dfrac{\left(1-x\right)\left(x+1\right)}{\left(x-1\right)\left(x-4\right)}=lim_{x\rightarrow1}\dfrac{-\left(x+1\right)}{x-4}=\dfrac{-\left(1+1\right)}{1-4}=\dfrac{2}{3}\)
d. Câu này mình chịu, nhìn đề hơi lạ so với bình thường hehe