Đáp án A, khi \(x\rightarrow1\) thì \(x-2< 0\) nên biểu thức không xác định

\(\Rightarrow\) Giới hạn đã cho ko tồn tại

Mình ko thấy đề bài

Bài 1:

\(a=\lim\limits_{x\rightarrow+\infty}\frac{\frac{1}{x}+\frac{2}{\sqrt{x}}-1}{1+\frac{3}{x}}=-1\)

\(b=\lim\limits_{x\rightarrow+\infty}\frac{1+\frac{3}{x^2}-\frac{1}{x^3}}{\frac{1}{\sqrt{x}}+\frac{1}{x^2}}=\frac{1}{0}=+\infty\)

\(c=\lim\limits_{x\rightarrow-\infty}\frac{1-2\sqrt{\frac{1}{x^2}-\frac{1}{x}}}{\frac{1}{x}-1}=\frac{1}{-1}=-1\)

Bài 2:

\(a=\lim\limits_{x\rightarrow0}\frac{1-cosx}{1-cos3x}=\lim\limits_{x\rightarrow0}\frac{sinx}{3sin3x}=\lim\limits_{x\rightarrow0}\frac{\frac{sinx}{x}}{9.\frac{sin3x}{3x}}=\frac{1}{9}\)

\(b=\lim\limits_{x\rightarrow0}\frac{cotx-sinx}{x^3}=\frac{\infty}{0}=+\infty\)

\(c=\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}\)

Mà \(\left|sinx\right|\le1\Rightarrow\left|\frac{sinx}{2x}\right|\le\frac{1}{\left|2x\right|}\)

Mà \(\lim\limits_{x\rightarrow\infty}\frac{1}{2\left|x\right|}=0\Rightarrow\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}=0\)

\(a=\lim\limits_{x\rightarrow-3}\frac{x^2+2x-3}{x\left(x+3\right)\left(x-\sqrt{3-2x}\right)}=\lim\limits_{x\rightarrow-3}\frac{\left(x-1\right)\left(x+3\right)}{x\left(x+3\right)\left(x-\sqrt{3-2x}\right)}=\lim\limits_{x\rightarrow-3}\frac{x-1}{x\left(x-\sqrt{3-2x}\right)}=-\frac{2}{9}\)

\(b=\lim\limits_{x\rightarrow0}\frac{\sqrt{x+9}-3+\sqrt{x+16}-4}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{x}{\sqrt{x+9}+3}+\frac{x}{\sqrt{x+16}+4}}{x}=\lim\limits_{x\rightarrow0}\left(\frac{1}{\sqrt{x+9}+3}+\frac{1}{\sqrt{x+16}+4}\right)=\frac{7}{24}\)

\(c=\lim\limits_{x\rightarrow\frac{1}{2}}\frac{8x^2-1}{6x^2-5x+1}\) ko phải dạng vô định, đề bài là \(8x^2\) hay \(8x^3\) bạn?

\(d=\lim\limits_{x\rightarrow0}\frac{\left(\sqrt{x^2+1}-1\right)\left(\sqrt{x^2+1}+1\right)\left(4+\sqrt{x^2+16}\right)}{\left(4-\sqrt{x^2+16}\right)\left(4+\sqrt{x^2+16}\right)\left(\sqrt{x^2+1}+1\right)}\)

\(=\lim\limits_{x\rightarrow0}\frac{x^2\left(4+\sqrt{x^2+16}\right)}{-x^2\left(\sqrt{x^2+1}+1\right)}=\lim\limits_{x\rightarrow0}\frac{4+\sqrt{x^2+16}}{-\sqrt{x^2+1}-1}=\frac{8}{-2}=-4\)

\(\lim\limits_{x\rightarrow-\infty}\frac{17}{x^2+4}=\frac{17}{+\infty}=0\)

\(\lim\limits_{x\rightarrow-\infty}\frac{-2x^2+x-1}{x^2+3}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}-\frac{1}{x^2}}{1+\frac{3}{x^2}}=\frac{-2+0+0}{1+0}=-2\)

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

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:

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

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

\(a=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}{\left(x-1\right)\left(x^2+x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x+1\right)\left(x^2+1\right)}{x^2+x-1}=\frac{4}{1}=4\)

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

\(c=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)^2}{\left(x^2+1\right)\left(x^2-9\right)}=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x+3\right)}=\frac{0}{60}=0\)

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

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

\(f=\lim\limits_{x\rightarrow-2}\frac{\left(x+2\right)\left(x-2\right)\left(x^2+4\right)}{\left(x+2\right)x^2}=\lim\limits_{x\rightarrow-2}\frac{\left(x-2\right)\left(x^2+4\right)}{x^2}=-8\)

Hai câu d, e khai triển thì dài quá nên làm biếng sử dụng L'Hopital

a/ Do \(x\rightarrow-3^+\) nên \(x>-3\Rightarrow x+3>0\Rightarrow\left|x+3\right|=x+3\)

\(\Rightarrow\lim\limits_{x\rightarrow-3^+}\frac{3x+9}{\left|x+3\right|}=\lim\limits_{x\rightarrow-3^+}\frac{3\left(x+3\right)}{x+3}=3\)

b/ \(=\lim\limits_{x\rightarrow0^+}\frac{\sqrt{x}\left(1-3\sqrt{x}\right)}{\sqrt{x}\left(4\sqrt{x}-2\right)}=\lim\limits_{x\rightarrow0^+}\frac{1-3\sqrt{x}}{4\sqrt{x}-2}=-\frac{1}{2}\)

Ở câu này \(x\rightarrow0^+\) có nghĩa \(x>0\), nó chỉ để căn thức xác định, ngoài ra ko có gì đặc biệt hết

c/ Tương tự câu c, cũng chỉ để căn thức xác định \(\left(x< 1\right)\)

\(\lim\limits_{x\rightarrow1^-}\frac{\sqrt{1-x}}{\left(1-x\right)\left(x+4\right)}=\lim\limits_{x\rightarrow1^-}\frac{1}{\sqrt{1-x}\left(x+4\right)}=+\infty\)

d/ Chắc bạn ghi nhầm đề, đây ko phải giới hạn dạng vô định (vì tử khác 0, mẫu bằng 0):

\(x\rightarrow\sqrt{2}^-\Rightarrow x< \sqrt{2}\Rightarrow x^4-4< 0\)

\(\Rightarrow\lim\limits_{x\rightarrow\sqrt{2}^-}\frac{\left|x-2\right|}{x^4-4}=-\infty\)

Xét giới hạn :

\(L=\lim\limits_{x\rightarrow\frac{\pi}{4}}\frac{1-\tan x}{1-\cot x}=\lim\limits_{x\rightarrow\frac{\pi}{4}}\frac{1-\frac{\sin x}{\cos x}}{1-\frac{\cos x}{\sin x}}=\lim\limits_{x\rightarrow\frac{\pi}{4}}\frac{\left(\cos x-\sin x\right)\sin x}{\left(\sin x-\cos x\right)\cos x}\)

   \(=-\lim\limits_{x\rightarrow\frac{\pi}{4}}\tan x=-1\)