\(lim\dfrac{\left(n+2\right)^{50}\left(n-3\right)^{80}}{\left(2n-1\right)^{40}\left(3n-2\right)^{45}}=lim\dfrac{\left(1+\dfrac{2}{n^{50}}\right)\left(1-\dfrac{3}{n^{35}}\right)\left(n-3\right)^{45}}{\left(2-\dfrac{1}{n^{50}}\right)\left(3-\dfrac{2}{n^{45}}\right)}=+\infty\)
\(lim\dfrac{4^n}{2.3^n+4^n}=lim\dfrac{1}{2.\left(\dfrac{3}{4}\right)^n+1}=\dfrac{1}{0+1}=1\)
\(lim\dfrac{3^n-2.5^n}{7+3.5^n}=lim\dfrac{\left(\dfrac{3}{5}\right)^n-2}{\dfrac{7}{5^n}+3}=\dfrac{0-2}{0+3}=\dfrac{-2}{3}\)
\(lim\dfrac{4^n-5^n}{2^{2n}+3.5^{2n}}=lim\dfrac{\left(\dfrac{4}{25}\right)^n-\left(\dfrac{1}{5}\right)^n}{\left(\dfrac{2}{5}\right)^{2n}+3}=\dfrac{0-0}{0+3}=0\)
\(lim\dfrac{\left(-3\right)^n+5^n}{2.\left(-4\right)^n+5^n}=lim\dfrac{\left(\dfrac{-3}{5}\right)^n+1}{2.\left(-\dfrac{4}{5}\right)^n+1}=\dfrac{0+1}{0+1}=1\)
1.
Nhớ rằng \(\lim _{x\to \infty}\frac{1}{x}=0\) và \(\lim _{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}\) với \(g(x)\neq 0; \lim_{x\to a}g(x)\neq 0\)
Do đó:
\(\lim_{n\to \infty}\frac{(n+2)^{50}.(n-3)^{80}}{(2n-1)^{40}.(3n-2)^{45}}=\lim_{n\to \infty}\frac{n^{130}(\frac{n+2}{n})^{50}.(\frac{n-3}{n})^{80}}{n^{85}(\frac{2n-1}{n})^{40}.(\frac{3n-2}{n})^{45}}\)
\(=\lim_{n\to \infty}\frac{n^{45}(1+\frac{2}{n})^{50}(1-\frac{3}{n})^{80}}{(2-\frac{1}{n})^{40}.(3-\frac{2}{n})^{45}}\)
\(=\frac{\lim_{n\to \infty}[n^{45}(1+\frac{2}{n})^{50}(1-\frac{3}{n})^{80}]}{\lim_{n\to \infty}[(2-\frac{1}{n})^{40}.(3-\frac{2}{n})^{45}]}\)
\(=\frac{\lim_{n\to \infty}n^{45}.1^{50}.1^{80}}{2^{40}.3^{45}}=\frac{\infty}{2^{40}.3^{45}}=\infty\)
2)
\(\lim_{n\to \infty}\frac{4^n}{2.3^n+4^n}=\lim_{n\to \infty}\frac{1}{\frac{2.3^n+4^n}{4^n}}=\lim_{n\to\infty}\frac{1}{2.(\frac{3}{4})^n+1}\)
\(=\frac{1}{\lim_{n\to \infty}[2.(\frac{3}{4})^n+1]}=\frac{1}{2.0+1}=1\)
3)
\(\lim_{n\to \infty}\frac{3^n-2.5^n}{7+3.5^n}=\lim_{n\to \infty}\frac{(\frac{3}{5})^n-2}{\frac{7}{5^n}+3}\)
\(=\frac{\lim_{n\to \infty}[(\frac{3}{5})^n-2]}{\lim_{n\to \infty}[\frac{7}{5^n}+3]}=\frac{0-2}{0+3}=\frac{-2}{3}\)
