Đặt \(t=x-\dfrac{\pi}{4}\), khi đó:

\(\lim\limits_{x\rightarrow\dfrac{\pi}{4}}\dfrac{\sqrt{2}cosx-1}{\sqrt{2}sinx-1}=\lim\limits_{t\rightarrow0}\dfrac{\sqrt{2}cos\left(t+\dfrac{\pi}{4}\right)-1}{\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)-1}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{cost-sint-1}{cost+sint-1}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{1-2sin^2\dfrac{t}{2}-2sin\dfrac{t}{2}.cos\dfrac{t}{2}-1}{1-2sin^2\dfrac{t}{2}+2sin\dfrac{t}{2}.cos\dfrac{t}{2}-1}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{-2sin\dfrac{t}{2}\left(sin\dfrac{t}{2}+cos\dfrac{t}{2}\right)}{-2sin\dfrac{t}{2}\left(sin\dfrac{t}{2}-cos\dfrac{t}{2}\right)}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{sin\dfrac{t}{2}+cos\dfrac{t}{2}}{sin\dfrac{t}{2}-cos\dfrac{t}{2}}\)

\(=-1\)

L'Hospital đi em

lim đề bài cho = \(\lim\limits_{x->\dfrac{\pi}{4}}\dfrac{-\sqrt{2}sinx}{\sqrt{2}cosx}\)

Thay x vào là xong

Tôi chẳng thể hiểu nổi

a: \(sinx=sin\left(\dfrac{\Omega}{4}\right)\)

=>\(\left[{}\begin{matrix}x=\dfrac{\Omega}{4}+k2\Omega\\x=\Omega-\dfrac{\Omega}{4}+k2\Omega=\dfrac{3}{4}\Omega+k2\Omega\end{matrix}\right.\)

b: cos2x=cosx

=>\(\left[{}\begin{matrix}2x=x+k2\Omega\\2x=-x+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k2\Omega\\3x=k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=k2\Omega\\x=\dfrac{k2\Omega}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{k2\Omega}{3}\)

c:

ĐKXĐ: \(x-\dfrac{\Omega}{3}< >\dfrac{\Omega}{2}+k\Omega\)

=>\(x< >\dfrac{5}{6}\Omega+k\Omega\)

 \(tan\left(x-\dfrac{\Omega}{3}\right)=\sqrt{3}\)

=>\(x-\dfrac{\Omega}{3}=\dfrac{\Omega}{3}+k\Omega\)

=>\(x=\dfrac{2}{3}\Omega+k\Omega\)

d:

ĐKXĐ: \(2x+\dfrac{\Omega}{6}< >k\Omega\)

=>\(2x< >-\dfrac{\Omega}{6}+k\Omega\)

=>\(x< >-\dfrac{1}{12}\Omega+\dfrac{k\Omega}{2}\)

 \(cot\left(2x+\dfrac{\Omega}{6}\right)=cot\left(\dfrac{\Omega}{4}\right)\)

=>\(2x+\dfrac{\Omega}{6}=\dfrac{\Omega}{4}+k\Omega\)

=>\(2x=\dfrac{1}{12}\Omega+k\Omega\)

=>\(x=\dfrac{1}{24}\Omega+\dfrac{k\Omega}{2}\)

\(1,\lim\limits_{n\rightarrow\infty}\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}\left(1\right)\)

\(\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}=\dfrac{-\dfrac{n^2}{n^4}+\dfrac{2n}{n^4}+\dfrac{1}{n^4}}{\sqrt{\dfrac{3n^4}{n^4}+\dfrac{2}{n^4}}}=\dfrac{-\dfrac{1}{n^2}+\dfrac{2}{n^3}+\dfrac{1}{n^4}}{\sqrt{3+\dfrac{2}{n^4}}}\)

\(\Rightarrow\left(1\right)=\dfrac{-lim\dfrac{1}{n^2}+2lim\dfrac{1}{n^3}+lim\dfrac{1}{n^4}}{\sqrt{lim\left(3+\dfrac{2}{n^4}\right)}}\)

\(=\dfrac{0}{\sqrt{lim\left(3+\dfrac{2}{n^4}\right)}}=0\)

\(2,\lim\limits_{n\rightarrow\infty}\left(\dfrac{4n-\sqrt{16n^2+1}}{n+1}\right)\left(2\right)\)

\(\dfrac{4n-\sqrt{16n^2+1}}{n+1}=\dfrac{\dfrac{4n}{n^2}-\sqrt{\dfrac{16n^2}{n^2}+\dfrac{1}{n^2}}}{\dfrac{n}{n^2}+\dfrac{1}{n^2}}=\dfrac{\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}}{\dfrac{1}{n}+\dfrac{1}{n^2}}\)

\(\Rightarrow\left(2\right)=\dfrac{lim\left(\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}\right)}{lim\left(\dfrac{1}{n}+\dfrac{1}{n^2}\right)}=\dfrac{lim\left(\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}\right)}{0}\)

Vậy giới hạn \(\left(2\right)\) không xác định.

\(3,\lim\limits_{n\rightarrow\infty}\left(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}\right)\left(3\right)\)

\(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}=\dfrac{\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}}{\dfrac{2}{n}}\)

\(\Rightarrow\left(3\right)=\dfrac{lim\left(\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}\right)}{2lim\dfrac{1}{n}}=\dfrac{lim\left(\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}\right)}{0}\)

Vậy \(lim\left(3\right)\) không xác định.

a: \(\lim\limits_{x->0^-^-}\dfrac{-2x+x}{x\left(x-1\right)}=lim_{x->0^-}\left(\dfrac{-x}{x\left(x-1\right)}\right)\)

\(=lim_{x->0^-}\left(\dfrac{-1}{x-1}\right)=\dfrac{-1}{0-1}=\dfrac{-1}{-1}=1\)

b: \(=lim_{x->-\infty}\left(\dfrac{x^2-x-x^2+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)

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

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