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

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

Câu b bạn coi lại đề, là \(x\rightarrow-1^-\) hay \(x\rightarrow1^-\) (đúng như đề thì ko phải dạng vô định, cứ thay số rồi bấm máy)

\(c=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)}{\left(x-3\right)\left(x-1\right)\left(\sqrt[3]{\left(x+5\right)^2}+2\sqrt[3]{x+5}+4\right)}\)

 \(=\lim\limits_{x\rightarrow3}\dfrac{1}{\left(x-1\right)\left(\sqrt[3]{\left(x+5\right)^2}+2\sqrt[3]{x+5}+4\right)}=\dfrac{1}{2.\left(4+4+4\right)}=...\)

a/ \(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}{\left(x-1\right)\left(x-3\right)}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1+\sqrt{3}\right)\left(x-1-\sqrt{3}\right)}{x-3}=....\)

Từ 2 câu kia lát tui làm, ăn cơm đã :D

\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-2x+1}{x-1}=3\rightarrow\lim\limits_{x\rightarrow1}\left(f\left(x\right)-2x+1\right)=0\\ \rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=1\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{3f\left(x\right)+1}-x-1}{\sqrt{4x+5}-3x-2}=\dfrac{\sqrt{3.1+1}-1-1}{\sqrt{4.1+5}-3.1-2}=0\)

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 1:

a. \(\lim\limits_{x\rightarrow-1}\frac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\frac{5x^4}{3x^2}=\frac{5}{3}\)

b. \(\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=\frac{120-100}{2}=10\)

c. \(\lim\limits_{x\rightarrow0}\frac{\left(1+2x\right)\left(1+3x\right)x}{x}+\lim\limits_{x\rightarrow0}\frac{\left(1+3x\right)2x}{x}+\lim\limits_{x\rightarrow0}\frac{3x+1-1}{x}=1+2+3=6\)

d. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)^5-\left(1+5x\right)}{x^5+x^2}=\lim\limits_{x\rightarrow0}\frac{5\left(1+x\right)^4-5}{5x^4+2x}\)

\(=\lim\limits_{x\rightarrow0}\frac{20\left(1+x\right)^3}{20x^3+2}=\frac{20}{2}=10\)

Bài 2:

\(\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)

\(\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}=\lim\limits_{x\rightarrow a}\frac{1}{nx^{n-1}}=\frac{1}{n.a^{n-1}}\)

1/ \(\lim\limits_{x\rightarrow0^-}\left(\dfrac{x-2}{x^3}\right)=\lim\limits_{x\rightarrow0^-}\dfrac{2-x}{-x^3}=\dfrac{2}{0}=+\infty\)

2/ \(\lim\limits_{x\rightarrow1^+}\dfrac{\left(x^3-x^2\right)^{\dfrac{1}{2}}}{\left(x-1\right)^{\dfrac{1}{2}}+1-x}=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{1}{2}\left(x^3-x^2\right)^{-\dfrac{1}{2}}.\left(3x^2-2x\right)}{\dfrac{1}{2}\left(x-1\right)^{-\dfrac{1}{2}}-1}=0\)

3/ \(\lim\limits_{x\rightarrow1^+}\dfrac{1-\left(x^2+x+1\right)}{x^3-1}=\dfrac{1-3}{0}=-\infty\)

4/ \(\lim\limits_{x\rightarrow-\infty}\left(-\infty-\sqrt[3]{1+\infty}\right)=-\left(\infty+\infty\right)=-\infty?\) Cái này ko chắc :v

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{4x+5}+x}{x^2+3x+2}=\dfrac{\sqrt{4.1+5}+1}{1^2+3.1+2}=\dfrac{2}{3}\)

(Đề là \(x\rightarrow-1\) thì hợp lý hơn)

\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{4x+5}+x}{x^2+3x+2}=\lim\limits_{x\rightarrow-1}\dfrac{\left(\sqrt{4x+5}+x\right)\left(\sqrt{4x+5}-x\right)}{\left(x+1\right)\left(x+2\right)\left(\sqrt{4x+5}-x\right)}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{-x^2+4x+5}{\left(x+1\right)\left(x+2\right)\left(\sqrt{4x+5}-x\right)}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(5-x\right)}{\left(x+1\right)\left(x+2\right)\left(\sqrt{4x+5}-x\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{5-x}{\left(x+2\right)\left(\sqrt{4x+5}-x\right)}=\dfrac{6}{1\left(1+1\right)}=3\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x-1-\sqrt{4x-3}\right)+\left(\sqrt[3]{6x-5}-\left(2x-1\right)\right)}{\left(x-1\right)^2}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{4\left(x-1\right)^2}{2x-1+\sqrt[]{4x-3}}-\dfrac{4\left(x-1\right)^2\left(2x+1\right)}{\sqrt[3]{\left(6x-5\right)^2}+\left(2x-1\right)\sqrt[3]{6x-5}+\left(2x-1\right)^2}}{\left(x-1\right)^2}\)

\(=\lim\limits_{x\rightarrow1}\left(\dfrac{4}{2x-1+\sqrt[]{4x-3}}-\dfrac{4\left(2x+1\right)}{\sqrt[3]{\left(6x-5\right)^2}+\left(2x-1\right)\sqrt[3]{6x-5}+\left(2x-1\right)^2}\right)=-2\)

\(lim\left(x->1\right)\dfrac{-\sqrt{4x-3}+\sqrt[3]{6x-5}}{\left(x-1\right)^2}\)

Đặt \(\sqrt{4x-3}=f\left(x\right);\sqrt[3]{6x-5}=g\left(x\right)\Rightarrow g\left(x\right)^6-f\left(x\right)^6=4\left(x-1\right)^2\left(16x-13\right)\)

\(f\left(1\right)=1;g\left(1\right)=1\)

Ta có 

\(lim\left(x->1\right)\dfrac{-f\left(x\right)+g\left(x\right)}{\left(x-1\right)^2}=lim\left(x->1\right)\dfrac{g\left(x\right)^6-f\left(x\right)^6}{\left(x-1\right)^2}\cdot\dfrac{1}{g\left(x\right)^5+g\left(x\right)^4f\left(x\right)+g\left(x\right)^3f\left(x\right)^2+g\left(x\right)^2f\left(x\right)^3+g\left(x\right)f\left(x\right)^4+f\left(x\right)^5}\)

\(=lim\left(x->1\right)\dfrac{4\left(16x-3\right)}{g\left(x\right)^5+g\left(x\right)^4f\left(x\right)+g\left(x\right)^3f\left(x\right)^2+g\left(x\right)^2f\left(x\right)^3+g\left(x\right)f\left(x\right)^4+f\left(x\right)^5}\)

\(=\dfrac{4\left(16-3\right)}{1^5+1^4\cdot1+1^3\cdot1^2+1^2\cdot1^3+1\cdot1^4+1^5}=\dfrac{26}{3}\)