\(\left\{{}\begin{matrix}-1\le sinx\le1\\-3\le3cos2x\le3\end{matrix}\right.\) \(\Rightarrow-4\le sinx+3cos3x\le4\) (dấu = có xảy ra hay ko ko hề quan trọng)
\(\Rightarrow\frac{-4}{x^2-2x+3}\le\frac{sinx+3cos2x}{x^2-2x+3}\le\frac{4}{x^2-2x+3}\)
Mà \(\lim\limits_{x\rightarrow\infty}\frac{-4}{x^2-2x+3}=\lim\limits_{x\rightarrow\infty}\frac{4}{x^2-2x+3}=0\)
\(\Rightarrow\lim\limits_{x\rightarrow\infty}\frac{sinx+3cos2x}{x^2-2x+3}=0\)
Tính các giới hạn
a) \(\lim\limits_{x\rightarrow+\infty}\sqrt[3]{x^3+3x^2}-\sqrt{x^2-2x}\)
b) \(\lim\limits_{x\rightarrow+\infty}\sqrt[n]{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)}-x\)
Tính giới hạn
a) \(\lim\limits_{x\rightarrow4^-}\dfrac{2x-5}{x-4}=-\infty\)
b) \(\lim\limits_{x\rightarrow+\infty}\left(-x^3+x^2-2x+1\right)\)
7)Tính giới hạn:
\(a)\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{x^2+2x}-2\sqrt{x^2+x}+x\right)\)
\(b)\lim\limits_{x\rightarrow+\infty}\left(\sqrt[3]{x^3+3x^2}-\sqrt{x^2-2x}\right)\)
Tìm giới hạn:
a, \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+x+2}}{x-1}\)
b, \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{4x^2-x}+2x\right)\)
\\(\\lim\\limits_{x\\rightarrow-\\infty}\\left(2x^3-x^2+3x-5\\right)\\)
\n\n\\(\\lim\\limits_{x\\rightarrow2}\\frac{3}{\\left(x-2\\right)\\left(x^2-3x+2\\right)}\\)
\n\n\\(\\lim\\limits_{x\\rightarrow0}\\frac{x^2-5}{x^5+x^4}\\)
\nTính các giới hạn
a) \(\lim\limits_{x\rightarrow4^-}\dfrac{2x-5}{x-4}\)
b) \(\lim\limits_{x\rightarrow+\infty}\left(-x^3+x^2-2x+1\right)\)
\(\lim\limits_{x\rightarrow\infty}\left(\sin\left(\sqrt{x+1}\right)-\sin\sqrt{x}\right)\)
Tính \(\lim\limits_{x\rightarrow+\infty}\left(\frac{2x}{x^2+x+1}+\frac{2x}{x^2+x+2}+...+\frac{2x}{x^2+2x}\right)\)
tính \(\lim\limits_{x\rightarrow+\infty}\left(x+1\right)\sqrt{\frac{2x+1}{x^3+x+2}}\)
