\(lim_{x->\pm\infty}\sqrt{x^2-3x+4}\)
\(lim_{x->\pm\infty}x\left(\sqrt{x^2+5}+x\right)\)
\(lim_{x->2019}\frac{\sqrt{x+285}-48}{\sqrt{x-2018}-\sqrt{2020-x}}\)
tìm các giới hạn sau
a)\(lim_{x->-\infty}\left(3x+\sqrt{1-2x+9x^2}\right)\)
b)\(lim_{x->+\infty\left(x-\sqrt{1+x+x^2}\right)}\)
\(lim_{x->+\infty}\left(\sqrt{x^2+2x}-2\sqrt{x^2+x}+x\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+2x}-\sqrt{x^2+x}+x-\sqrt{x^2+x}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{x}{\sqrt{x^2+2x}+\sqrt{x^2+x}}-\dfrac{x}{x+\sqrt{x^2+x}}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{1}{\sqrt{1+\dfrac{2}{x}}+\sqrt{1+\dfrac{1}{x}}}-\dfrac{1}{1+\sqrt{1+\dfrac{1}{x}}}\right)=\dfrac{1}{2}-\dfrac{1}{2}=0\)
\(lim_{x\rightarrow+\infty}\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\frac{x+\sqrt{x+\sqrt{x}}-x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\)
\(=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\)
\(=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\sqrt{\frac{1}{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x^3}}}}+1}=\frac{1}{1+1}=\frac{1}{2}\)
\(lim_{x->\frac{+}{ }\infty}\frac{\sqrt{x^2+3x+5}}{\sqrt[3]{x^3+7x^2+8}}\)
Giới hạn này tiến đến đâu vậy bạn? 2 trường hợp khác nhau đúng ko?
\(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x^2+3x+5}}{\sqrt[3]{x^3+7x^2+8}}=\lim\limits_{x\rightarrow+\infty}\frac{x\sqrt{1+\frac{3}{x}+\frac{5}{x^2}}}{x\sqrt[3]{1+\frac{7}{x}+\frac{8}{x^3}}}=1\)
\(\lim\limits_{x\rightarrow-\infty}\frac{\sqrt{x^2+3x+5}}{\sqrt[3]{x^3+7x^2+8}}=\lim\limits_{x\rightarrow-\infty}\frac{\left|x\right|\sqrt{1+\frac{3}{x}+\frac{5}{x^2}}}{x\sqrt[3]{1+\frac{7}{x}+\frac{8}{x^3}}}=\lim\limits_{x\rightarrow-\infty}\frac{-x\sqrt{1+\frac{3}{x}+\frac{5}{x^2}}}{x\sqrt[3]{1+\frac{7}{x}+\frac{8}{x^3}}}=-1\)
\(\lim_{x\to -\infty} ((2x+1)^2+4\sqrt{x^2+4}\sqrt[3]{x^3+3x^2})\)
\(lim_{x->-\infty}\left(\sqrt{x^2+1}+x-1\right)\\ lim\dfrac{\sqrt{4n^2+n-1}+n}{\sqrt{n^4+2n^3-1}-n}\)
\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2+1}+x-1\right)\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+1-\left(x-1\right)^2}{\sqrt{x^2+1}-x+1}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+1-x^2+2x-1}{-x\sqrt{1+\dfrac{1}{x^2}}-x+1}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-2x}{x\left(-\sqrt{1+\dfrac{1}{x^2}}-1+\dfrac{1}{x}\right)}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-2}{-\sqrt{1+\dfrac{1}{x^2}}-1+\dfrac{1}{x}}\)
\(=\dfrac{-2}{-\sqrt{1+0}-1+0}=\dfrac{-2}{-1-1}=1\)
b: \(\lim\limits\dfrac{\sqrt{4n^2+n-1}+n}{\sqrt{n^4+2n^3-1}-n}\)
\(=\lim\limits\dfrac{n\left(\sqrt{4+\dfrac{1}{n}-\dfrac{1}{n^2}}+1\right)}{n^2\cdot\sqrt{1+\dfrac{2}{n}-\dfrac{1}{n^4}}-n^2\cdot\dfrac{1}{n}}\)
\(=\lim\limits\dfrac{n\left(\sqrt{4+\dfrac{1}{n}-\dfrac{1}{n^2}}+1\right)}{n^2\left(\sqrt{1+\dfrac{2}{n}-\dfrac{1}{n^4}}-\dfrac{1}{n}\right)}\)
\(=\lim\limits\dfrac{\sqrt{4+\dfrac{1}{n}-\dfrac{1}{n^2}}+1}{n\left(\sqrt{1+\dfrac{2}{n}-\dfrac{1}{n^4}}-\dfrac{1}{n}\right)}\)
\(=\lim\limits\dfrac{\sqrt{\dfrac{4}{n^2}+\dfrac{1}{n^3}-\dfrac{1}{n^4}}+\dfrac{1}{n}}{\sqrt{1+\dfrac{2}{n}-\dfrac{1}{n^4}}-\dfrac{1}{n}}\)
\(=\dfrac{0}{\sqrt{1+0-0}-0}=\dfrac{0}{1}=0\)
tìm a để giới hạn : \(lim_{x->+\infty}\left(\sqrt{x^2-3x+5}+ax\right)=+\infty\)
Bạn tự hiểu là giới hạn khi x tiến tới dương vô cực
\(=lim\left[x\left(\sqrt{1-\frac{3}{x}+\frac{5}{x^2}}+a\right)\right]=lim\left[x\left(1-a\right)\right]\)
Do \(x\rightarrow+\infty\) nên để giới hạn đã cho bằng \(+\infty\Leftrightarrow1-a>0\Rightarrow a< 1\)
Tính giới hạn
a, \(Lim_{n->+\infty}\frac{1+sin\left(n\right)+2^{n+2}}{2-2n+2^n}\)
b,\(Lim_{x->0}\frac{e^x-1-xcos\left(x\right)}{x\left(e^{2x}-1\right)}\)
c,\(Lim_{n->+\infty}\sqrt[2n]{8^n+9^n}\)
d,\(Lim_{x->0}\frac{\ln\left(1+x\right)-xe^3}{x\tan\left(2x\right)}\)
Tính giới hạn
a, \(Lim_{n->+\infty}\frac{1+sin\left(n\right)+2^{n+2}}{2-2n+2^n}\)
b,\(Lim_{x->0}\frac{e^x-1-xcos\left(x\right)}{x\left(e^{2x}-1\right)}\)
c,\(Lim_{n->+\infty}\sqrt[2n]{8^n+9^n}\)
d,\(Lim_{x->0}\frac{\ln\left(1+x\right)-xe^3}{x\tan\left(2x\right)}\)