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

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

Mấy câu này bạn cần giải theo kiểu trắc nghiệm hay tự luận nhỉ?

Em cần kiểu tự luận ạ

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

\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-16}{x-1}\) hữu hạn nên \(f\left(x\right)-16=0\) có nghiệm \(x=1\)

\(\Rightarrow f\left(1\right)=16\)

\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-16}{x-1}.\dfrac{1}{\sqrt{2f\left(x\right)+4}+6}=24.\dfrac{1}{\sqrt{2.16+4}+6}=2\)

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

\(a=\lim\limits_{x\rightarrow a}\frac{\left(\sqrt{x}-\sqrt{a}\right)\left(x+\sqrt{ax}+a\right)}{\sqrt{x}-\sqrt{a}}=\lim\limits_{x\rightarrow a}\left(x+\sqrt{ax}+a\right)=3a\)

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

Ta có:

\(\lim\limits_{x\rightarrow1}\frac{1-\sqrt[n]{x}}{1-x}=\lim\limits_{x\rightarrow1}\frac{1-x^{\frac{1}{n}}}{1-x}=\lim\limits_{x\rightarrow1}\frac{-\frac{1}{n}x^{\frac{1-n}{n}}}{-1}=\frac{1}{n}\)

\(\Rightarrow c=\lim\limits_{x\rightarrow1}\frac{\left(1-\sqrt{x}\right)}{1-x}.\frac{\left(1-\sqrt[3]{x}\right)}{\left(1-x\right)}.\frac{\left(1-\sqrt[4]{x}\right)}{\left(1-x\right)}.\frac{\left(1-\sqrt[5]{x}\right)}{\left(1-x\right)}=\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}=\frac{1}{120}\)

\(d=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x\sqrt{x}}}}+1}=\frac{1}{2}\)

\(e=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+x}-1+1-\sqrt[3]{1+x}}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{x}{\sqrt{1+x}+1}+\frac{x}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\frac{1}{\sqrt{1+x}+1}+\frac{1}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}\right)=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

\(f=\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-3+3-\sqrt{x+7}}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2}\frac{\frac{8\left(x-2\right)}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{x-2}{3+\sqrt{x+7}}}{\left(x-1\right)\left(x-2\right)}\)

\(=\lim\limits_{x\rightarrow2}\frac{\frac{8}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{1}{3+\sqrt{x+7}}}{x-1}=\frac{8}{27}-\frac{1}{6}=\frac{7}{54}\)

\(g=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{3x-2}-1+1-\sqrt{2x-1}}{\left(x-1\right)\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{3\left(x-1\right)}{\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1}-\frac{2\left(x-1\right)}{1+\sqrt{2x-1}}}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\frac{\frac{3}{\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1}-\frac{2}{1+\sqrt{2x-1}}}{x^2+x+1}=0\)

\(h=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+9}+\sqrt[3]{2x-6}}{x^3+1}=\frac{\sqrt[3]{10}-\sqrt[3]{4}}{2}\)

\(a=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-1+1-\sqrt[3]{2x+1}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x}{\sqrt[]{4x+1}+1}+\dfrac{-2x}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{4x+1}+1}+\dfrac{-2}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}\right)=...\)

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

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

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

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

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)

\(I_2=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+x^2}+x\right)^n-\left(\sqrt{1+x^2}-x\right)^n}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{n\left(\sqrt{1+x^2}+x\right)^{n-1}\left(\dfrac{x}{\sqrt{1+x^2}}+1\right)-n\left(\sqrt{1+x^2}-x\right)^{n-1}\left(\dfrac{x}{\sqrt{1+x^2}}-1\right)}{1}\)

\(=\dfrac{n.1-n\left(-1\right)}{1}=2n\)