Chia cả tử và mẫu cho x2, lim đề bài cho sẽ bằng với
\(lim\dfrac{\sqrt{1-\dfrac{1}{n^2}}+3}{\dfrac{1}{n^2}-1}=\dfrac{\sqrt{1-0}+3}{0-1}=-4\)
\(=\lim\dfrac{\sqrt{1-\dfrac{1}{n^2}}+3}{\dfrac{1}{n^2}-1}=\dfrac{1+3}{-1}=-4\)
Chia cả tử và mẫu cho x2, lim đề bài cho sẽ bằng với
\(lim\dfrac{\sqrt{1-\dfrac{1}{n^2}}+3}{\dfrac{1}{n^2}-1}=\dfrac{\sqrt{1-0}+3}{0-1}=-4\)
\(=\lim\dfrac{\sqrt{1-\dfrac{1}{n^2}}+3}{\dfrac{1}{n^2}-1}=\dfrac{1+3}{-1}=-4\)
1/ lim \(\dfrac{\sqrt{n^4-n^2}+3n^2}{1-n^2}\)
2/ lim \(\dfrac{n\sqrt{n}-n^3}{4n^3+\sqrt{n}}\)
3/ lim \(\dfrac{3.4^n-1}{2.3^n+4}\)
4/ lim \(\dfrac{2^{n+1}+4.3^{n-1}}{1-2^{n-1}+3^{n+1}}\)
1) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}\)
2) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\dfrac{4n-\sqrt{16n^2+1}}{n+1}\right)\)
3) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}\right)\)
tính
1) \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5n-3}{-n+5}\)
2) \(\lim\limits_{n\rightarrow\infty}\dfrac{-7n^2+4}{-n+5}\)
3) \(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^2+2}{n-2}\)
1) \(\lim\limits_{n\rightarrow\infty}\dfrac{-7n^2+4}{-n+5}\)
2) \(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^2+2}{n-2}\)
1) tính \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4x^4-3n^2+4\right)\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^2+2}{n-2}\)
1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^3+3n^2-1}{n^2-2n}\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2-1}{-2n+3}\)
1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5n-3}{-n+5}\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{-7n^2+4}{-n+5}\)
1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^5+3n^3-1}{n^3-2n}\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^7+3n^5-n}{3n^2-2n}\)
1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}\)
2) \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}\)
3) \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4x^4-3n^2+4\right)\)