\(=lim\frac{2\left(2^n-1\right)}{\frac{5\left(5^n-1\right)}{4}}=\frac{8.2^n-8}{5.5^n-5}=\frac{8\left(\frac{2}{5}\right)^n-8\left(\frac{1}{5}\right)^n}{5-5\left(\frac{1}{5}\right)^n}=\frac{0}{5}=0\)
\(=lim\frac{2\left(2^n-1\right)}{\frac{5\left(5^n-1\right)}{4}}=\frac{8.2^n-8}{5.5^n-5}=\frac{8\left(\frac{2}{5}\right)^n-8\left(\frac{1}{5}\right)^n}{5-5\left(\frac{1}{5}\right)^n}=\frac{0}{5}=0\)
lim \(\frac{1}{\sqrt[4]{64n^4+3n^3-2n^2+1}-\sqrt{n^2-3n+5}-3n}\)
1. tính gới hạn \(\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+7}-\sqrt{5-x^2}}{x-1}\)
2. tính gới hạn của dãy số \(\lim\limits\frac{1^7+2^7+...+n^7}{n^8}\)
giúp e tự luận với ah
\(\lim\limits_{ }\left(\frac{2}{7}+\frac{16}{49}+...+\frac{5^n-3^n}{7^n}\right)\)
Tính 1) \(lim\frac{\sqrt{n}-2}{n+\sqrt{n}+1}\)
2) \(lim\frac{\sqrt[3]{n^3+n}+2}{n+2}\)
3)\(lim\frac{\sqrt[3]{n^3+1}-1}{\sqrt{n^2+3}-2}\)
Tính các giới hạn sau (\(n\rightarrow+\infty\) )
a) \(\lim\limits\dfrac{\left(-3\right)^n+2.5^n}{1-5^n}\)
b) \(\lim\limits\dfrac{1+2+3+....+n}{n^2+n+1}\)
c) \(\lim\limits\left(\sqrt{n^2+2n+1}-\sqrt{n^2+n-1}\right)\)
\(\lim\limits_{\rightarrow2}\) \(\frac{\sqrt{x^2+5}-3}{2-x}\)
Tính giới hạn sau :
lim x tiến tới -2\(\frac{2\left|x-1\right|-5\sqrt{x^2-3}}{2x+3}\)
Tính :
\(\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{6x-5}-\sqrt{4x-3}}{\left(x-1\right)^2}\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2-x+1}-x\right)\)
\(\lim\limits_{x\rightarrow-\infty}x\left(\sqrt{4x^2+1}-x\right)\)
\(\lim\limits_{x\rightarrow-\infty}\left(4x^5-3x^3+x+1\right)\)
\(\lim\limits_{x\rightarrow+\infty}\sqrt{x^4-x^3+x^2-x}\)