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Em là mạnh Tuấn ạ

a: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x-2}-\dfrac{12}{x^3-8}\right)\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x+4-12}{\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x-8}{\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x+4}{x^2+2x+4}\)

\(=\dfrac{2+4}{2^2+2\cdot2+4}=\dfrac{6}{4+4+4}=\dfrac{6}{12}=\dfrac{1}{2}\)

b: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\)

\(=\lim\limits_{x\rightarrow2}\left(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}\right)\)

\(=\lim\limits_{x\rightarrow2}\left(\dfrac{x-3+x-1}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\right)\)

\(=\lim\limits_{x\rightarrow2}\dfrac{2x-4}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{2}{\left(x-3\right)\left(x-1\right)}=\dfrac{2}{\left(2-3\right)\left(2-1\right)}=-2\)

d: \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-\sqrt[3]{x^3-1}\right)\)

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

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

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

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

=0

c: \(\lim\limits_{x\rightarrow+\infty}\left[x\cdot\left(\sqrt{x^2+1}-x\right)\right]\)

\(=\lim\limits_{x\rightarrow+\infty}\left[x\cdot\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}\right]\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x}{\sqrt{x^2+1}+x}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}+1}=\dfrac{1}{1+1}=\dfrac{1}{2}\)

e: \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{x^2+1-1}{\sqrt{x^2+1}+1}:\dfrac{x^2+16-16}{\sqrt{x^2+16}+4}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+16}+4}{\sqrt{x^2+1}+1}=\dfrac{4+4}{1+1}=\dfrac{8}{2}=4\)

\(\lim\limits_{x\rightarrow6}\dfrac{f\left(x\right)-6}{x-6}\) hữu hạn \(\Rightarrow f\left(6\right)=6\)

\(...=\lim\limits_{x\rightarrow6}\dfrac{\dfrac{f\left(x\right)-6}{\sqrt[3]{\left[f\left(x\right)+21\right]^2}+3\sqrt[3]{f\left(x\right)+21}+9}}{x-6}\)

\(=\dfrac{9}{2}.\dfrac{1}{\sqrt[3]{\left(6+21\right)^2}+3\sqrt[3]{6+21}+9}\)

Do \(\lim\limits_{x\rightarrow-1}\dfrac{2f\left(x\right)+1}{x+1}=5\) hữu hạn nên \(2f\left(x\right)+1=0\) phải có nghiệm \(x=-1\)

\(\Leftrightarrow2f\left(-1\right)=-1\Leftrightarrow f\left(-1\right)=-\dfrac{1}{2}\)

Đoạn dưới tự hiểu là \(\lim\limits_{x\rightarrow-1}\) (vì kí tự lim rất rắc rối)

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

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

\(=\dfrac{2f\left(x\right)+1}{x+1}.\dfrac{f\left(x\right).\left[4f\left(x\right)+3\right]}{x-1}+\dfrac{2f\left(x\right)+1}{x+1}.\dfrac{4}{x-1}\)

\(=5.\dfrac{f\left(-1\right).\left[4f\left(-1\right)+3\right]}{-2}+5.\dfrac{4}{-2}=\dfrac{5.\left(-\dfrac{1}{2}\right)\left(-2+3\right)}{-2}+5.\dfrac{4}{-2}=...\)

1: \(A=\dfrac{x^2-\left(a+1\right)x+a}{x^3-a^3}\)

\(=\dfrac{x^2-xa-x+a}{\left(x-a\right)\left(x^2+ax+a^2\right)}\)

\(=\dfrac{\left(x-a\right)\left(x-1\right)}{\left(x-a\right)\left(x^2+ax+a^2\right)}=\dfrac{x-1}{x^2+ax+a^2}\)

\(lim_{x->a}A=lim_{x->a}\left(\dfrac{x-1}{x^2+ax+a^2}\right)\)

\(=\dfrac{a-1}{a^2+a^2+a^2}=\dfrac{a-1}{3a^2}\)

2: \(B=\dfrac{1}{1-x}-\dfrac{3}{1-x^3}\)

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

\(=\dfrac{-\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-x-2}{x^2+x+1}\)

\(lim_{x->1}\left(B\right)=\dfrac{-1-2}{1^2+1+1}=\dfrac{-3}{3}=-1\)

3: \(C=\dfrac{\left(x+h\right)^3-x^3}{h}=\dfrac{\left(x+h-x\right)\left(x^2+2xh+h^2+x^2+hx+x^2\right)}{h}\)

\(=3x^2+3hx\)

\(lim_{h->0}\left(C\right)=3x^2+3\cdot0\cdot x=3x^2\)

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

Chọn F(x)=5x-23

\(\lim\limits_{x\rightarrow5}\dfrac{f\left(x\right)-2}{x-5}=\lim\limits_{x\rightarrow5}\dfrac{5x-23-2}{x-5}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{5x-25}{x-5}=\lim\limits_{x\rightarrow5}\dfrac{5\left(x-5\right)}{x-5}=5\)

=>f(x)=5x-23 thỏa mãn yêu cầu đề bài

\(\lim\limits_{x\rightarrow5}\dfrac{\sqrt{3\cdot f\left(x\right)+10}+\sqrt{f^3\left(x\right)+1}-7}{x^2-25}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\sqrt{3\left(5x-23\right)+10}+\sqrt{\left(5x-23\right)^3+1}-7}{\left(x-5\right)\left(x+5\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\sqrt{15x-59}+\sqrt{\left(5x-23\right)^3+1}-7}{\left(x-5\right)\left(x+5\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\sqrt{15x-59}-4+\sqrt{\left(5x-23\right)^3+1}-3}{\left(x-5\right)\left(x+5\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{15x-59-16}{\sqrt{15x-59}+4}+\dfrac{\left(5x-23\right)^3+1-9}{\sqrt{\left(5x-23\right)^3+1}+3}}{\left(x-5\right)\left(x+5\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{15\left(x-5\right)}{\sqrt{15x-59}+4}+\dfrac{\left(5x-23\right)^3-8}{\sqrt{\left(5x-23\right)^3+1}+3}}{\left(x-5\right)\left(x+5\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{15\left(x-5\right)}{\sqrt{15x-59}+4}+\dfrac{\left(5x-23-2\right)\left[\left(5x-23\right)^2+2\left(5x-23\right)+4\right]}{\sqrt{\left(5x-23\right)^3+1}+3}}{\left(x-5\right)\left(x+5\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{15}{\sqrt{15x-59}+4}+\dfrac{5\cdot\left(25x^2-230x+529+10x-46+4\right)}{\sqrt{\left(5x-23\right)^3+1}+3}}{x+5}\)

\(=\dfrac{\dfrac{15}{\sqrt{15\cdot5-59}+4}+\dfrac{5\left(25\cdot5^2-220\cdot5+487\right)}{\sqrt{\left(5\cdot5-23\right)^3+1}+3}}{5+5}\)

\(=\dfrac{\dfrac{15}{8}+\dfrac{5\cdot12}{6}}{10}=\dfrac{19}{16}\)

Do \(\lim\limits_{x\rightarrow5}\dfrac{f\left(x\right)-2}{x-5}\) hữu hạn nên \(f\left(x\right)-2=0\) có nghiệm \(x=5\)

\(\Rightarrow f\left(5\right)=2\)

\(\lim\limits_{x\rightarrow5}\dfrac{\sqrt{3f\left(x\right)+10}-4+\sqrt{f^3\left(x\right)+1}-3}{\left(x-5\right)\left(x+5\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{3\left[f\left(x\right)-2\right]}{\sqrt{3f\left(x\right)+10}+4}+\dfrac{\left[f\left(x\right)-2\right]\left[f^2\left(x\right)+2f\left(x\right)+4\right]}{\sqrt{f^3\left(x\right)+1}+3}}{\left(x-5\right)\left(x+5\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{f\left(x\right)-2}{x-5}.\dfrac{3}{\sqrt{3f\left(x\right)+10}+4}+\dfrac{f\left(x\right)-2}{x-5}.\dfrac{f^2\left(x\right)+2f\left(x\right)+4}{\sqrt{f^3\left(x\right)+1}+3}}{x+5}\)

\(=\dfrac{5.\dfrac{3}{\sqrt{3.2+10}+4}+5.\dfrac{2^2+2.2+4}{\sqrt{2^3+1}+3}}{5+5}=\)

1.

Trước hết bạn nhớ công thức:

$1^2+2^2+....+n^2=\frac{n(n+1)(2n+1)}{6}$ (cách cm ở đây: https://hoc24.vn/cau-hoi/tinh-tongs-122232n2.83618073020)

Áp vào bài:

\(\lim\frac{1}{n^3}[1^2+2^2+....+(n-1)^2]=\lim \frac{1}{n^3}.\frac{(n-1)n(2n-1)}{6}=\lim \frac{n(n-1)(2n-1)}{6n^3}\)

\(=\lim \frac{(n-1)(2n-1)}{6n^2}=\lim (\frac{n-1}{n}.\frac{2n-1}{6n})=\lim (1-\frac{1}{n})(\frac{1}{3}-\frac{1}{6n})\)

\(=1.\frac{1}{3}=\frac{1}{3}\)

2.

\(\lim \frac{1}{n}\left[(x+\frac{a}{n})+(x+\frac{2a}{n})+...+(x.\frac{(n-1)a}{n}\right]\)

\(=\lim \frac{1}{n}\left[\underbrace{(x+x+...+x)}_{n-1}+\frac{a(1+2+...+n-1)}{n} \right]\)

\(=\lim \frac{1}{n}[(n-1)x+a(n-1)]=\lim \frac{n-1}{n}(x+a)=\lim (1-\frac{1}{n})(x+a)\)

\(=x+a\) 

3.

Trước tiên ta có công thức:

$1^3+2^3+....+n^3=(1+2+3+...+n)^2=\frac{n^2(n+1)^2}{4}$
Chứng minh: https://diendantoanhoc.org/topic/81694-t%C3%ADnh-t%E1%BB%95ng-s-13-23-33-n3/

Khi đó:

\(\lim \frac{1^3+2^3+...+n^3}{n^4}=\lim \frac{n^2(n+1)^2}{4n^4}\\ =\lim \frac{(n+1)^2}{4n^2}=\frac{1}{4}\lim (1+\frac{1}{n})^2=\frac{1}{4}.1=\frac{1}{4}\)

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)

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

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

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

Do \(\lim\limits_{x\rightarrow3}\dfrac{f\left(x\right)-2}{x-3}\) hữu hạn \(\Rightarrow f\left(x\right)-2=0\) có nghiệm \(x=3\)

Hay \(f\left(3\right)-2=0\Rightarrow f\left(3\right)=2\)

\(\Rightarrow I=\lim\limits_{x\rightarrow3}\left(\dfrac{f\left(x\right)-2}{x-3}\right).\dfrac{1}{\sqrt{5f\left(x\right)+6}+1}=\dfrac{1}{4}.\dfrac{1}{\sqrt{5.f\left(3\right)+6}+1}\)

\(=\dfrac{1}{4}.\dfrac{1}{\sqrt{5.2+6}+1}=\dfrac{1}{20}\)