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\(\lim\limits_{x\rightarrow1}\dfrac{x^3-3x+2}{x^4-4x+3}=\lim\limits_{x\rightarrow1}\dfrac{\left(x+2\right)\left(x-1\right)^2}{\left(x^2+2x+3\right)\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\dfrac{x+2}{x^2+2x+3}=\dfrac{1}{2}\)
\(\lim\limits_{x\rightarrow2^-}\dfrac{x^3+x^2-4x-4}{x^2-4x+4}=\lim\limits_{x\rightarrow2^-}\dfrac{\left(x-2\right)\left(x^2+3x+2\right)}{\left(x-2\right)^2}=\lim\limits_{x\rightarrow2^-}\dfrac{x^2+3x+2}{x-2}=-\infty\)
\(\lim\limits_{x\rightarrow2}\dfrac{\left(x^2-x-2\right)^{20}}{\left(x^3-12x+16\right)^{10}}=\lim\limits_{x\rightarrow2}\dfrac{\left(x+1\right)^{20}\left(x-2\right)^{20}}{\left(x+4\right)^{10}\left(x-2\right)^{20}}=\lim\limits_{x\rightarrow2}\dfrac{\left(x+1\right)^{20}}{\left(x+4\right)^{10}}=\dfrac{3^{10}}{2^{10}}\)
\(\lim\limits_{x\rightarrow0^-}\dfrac{4x^2+5x}{x^2}=\lim\limits_{x\rightarrow0^-}\dfrac{4x+5}{x}=-\infty\)
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{x+2}-1}{\sqrt{x+5}-2}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(\sqrt{x+5}+2\right)}{\left(x+1\right)\left(\sqrt{x+2}+1\right)}=\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{x+5}+2}{\sqrt{x+2}+1}=2\)
a) Ta có (x - 2)2 = 0 và (x - 2)2 > 0 với ∀x ≠ 2 và
(3x - 5) = 3.2 - 5 = 1 > 0.
Do đó
= +∞.
b) Ta có (x - 1) và x - 1 < 0 với ∀x < 1 và
(2x - 7) = 2.1 - 7 = -5 <0.
Do đó
= +∞.
c) Ta có (x - 1) = 0 và x - 1 > 0 với ∀x > 1 và
(2x - 7) = 2.1 - 7 = -5 < 0.
Do đó
= -∞.
\(lim\dfrac{\left(n+2\right)^{50}\left(n-3\right)^{80}}{\left(2n-1\right)^{40}\left(3n-2\right)^{45}}=lim\dfrac{\left(1+\dfrac{2}{n^{50}}\right)\left(1-\dfrac{3}{n^{35}}\right)\left(n-3\right)^{45}}{\left(2-\dfrac{1}{n^{50}}\right)\left(3-\dfrac{2}{n^{45}}\right)}=+\infty\)
\(lim\dfrac{4^n}{2.3^n+4^n}=lim\dfrac{1}{2.\left(\dfrac{3}{4}\right)^n+1}=\dfrac{1}{0+1}=1\)
\(lim\dfrac{3^n-2.5^n}{7+3.5^n}=lim\dfrac{\left(\dfrac{3}{5}\right)^n-2}{\dfrac{7}{5^n}+3}=\dfrac{0-2}{0+3}=\dfrac{-2}{3}\)
\(lim\dfrac{4^n-5^n}{2^{2n}+3.5^{2n}}=lim\dfrac{\left(\dfrac{4}{25}\right)^n-\left(\dfrac{1}{5}\right)^n}{\left(\dfrac{2}{5}\right)^{2n}+3}=\dfrac{0-0}{0+3}=0\)
\(lim\dfrac{\left(-3\right)^n+5^n}{2.\left(-4\right)^n+5^n}=lim\dfrac{\left(\dfrac{-3}{5}\right)^n+1}{2.\left(-\dfrac{4}{5}\right)^n+1}=\dfrac{0+1}{0+1}=1\)
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:
\(\lim_{n\to \infty}\frac{(n+2)^{50}.(n-3)^{80}}{(2n-1)^{40}.(3n-2)^{45}}=\lim_{n\to \infty}\frac{n^{130}(\frac{n+2}{n})^{50}.(\frac{n-3}{n})^{80}}{n^{85}(\frac{2n-1}{n})^{40}.(\frac{3n-2}{n})^{45}}\)
\(=\lim_{n\to \infty}\frac{n^{45}(1+\frac{2}{n})^{50}(1-\frac{3}{n})^{80}}{(2-\frac{1}{n})^{40}.(3-\frac{2}{n})^{45}}\)
\(=\frac{\lim_{n\to \infty}[n^{45}(1+\frac{2}{n})^{50}(1-\frac{3}{n})^{80}]}{\lim_{n\to \infty}[(2-\frac{1}{n})^{40}.(3-\frac{2}{n})^{45}]}\)
\(=\frac{\lim_{n\to \infty}n^{45}.1^{50}.1^{80}}{2^{40}.3^{45}}=\frac{\infty}{2^{40}.3^{45}}=\infty\)
