sao có GP lại ko có huy hiệu hỏi thôi

Thấy : \(\sqrt{x^2+x+3}-x^2+1=\sqrt{x^2+x+3}-\left(x^2-1\right)=\dfrac{x^2+x+3-\left(x^2-1\right)^2}{\sqrt{x^2+x+3}+x^2-1}\)

\(=\dfrac{x^2+x+3-x^4+2x^2-1}{...}=\dfrac{-x^4+3x^2+x+2}{...}\)

\(=\dfrac{-\left(x-2\right)\left(x^3+2x^2+x+1\right)}{...}\)

\(\dfrac{\sqrt{x^2+x+3}-x^2+1}{x^2-4}=\dfrac{-\left(x^3+2x^2+x+1\right)}{\left(x+2\right)\left[\sqrt{x^2+x+3}+x^2-1\right]}\)

\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{x^2+x+3}-x^2+1}{x^2-4}=\dfrac{-\left(2^3+2.2^2+2+1\right)}{4.\left[\sqrt{2^2+2+3}+2^2-1\right]}=-\dfrac{19}{24}\)

\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{x^2+x+3}-x^2+1}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{2x+1}{2\sqrt{x^2+x+3}}-2x}{2x}=\dfrac{\dfrac{2.2+1}{2\sqrt{4+2+3}}-4}{4}=-\dfrac{19}{24}\)

Lời giải:

a. \(\lim\limits_{x\to 1+}(x^3+x+1)=3>0\)

\(\lim\limits_{x\to 1+}(x-1)=0\) và $x-1>0$ khi $x>1$

\(\Rightarrow \lim\limits_{x\to 1+}\frac{x^3+x+1}{x-1}=+\infty\)

b.

 \(\lim\limits_{x\to -1+}(3x+2)=-1<0\)

\(\lim\limits_{x\to -1+}(x+1)=0\) và $x+1>0$ khi $x>-1$

\(\Rightarrow \lim\limits_{x\to -1+}\frac{3x+2}{x+1}=-\infty\)

c.

\(\lim\limits_{x\to 2-}(x-15)=-17<0\)

\(\lim\limits_{x\to 2-}(x-2)=0\) và $x-2<0$ khi $x<2$

\(\Rightarrow \lim\limits_{x\to 2-}\frac{x-15}{x-2}=+\infty\)

a/ \(\lim\limits_{x\rightarrow2}\dfrac{2+3}{4+2+4}=\dfrac{5}{10}=\dfrac{1}{2}\)

b/ \(\lim\limits_{x\rightarrow-3}\dfrac{\left(x+2\right)\left(x+3\right)}{x\left(x+3\right)}=\lim\limits_{x\rightarrow-3}\dfrac{x+2}{x}=\dfrac{-3+2}{-3}=\dfrac{1}{3}\)

a: \(\lim\limits_{x\rightarrow2}\dfrac{1-\sqrt{x^2+3}}{-x^2+3x-2}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\sqrt{x^2+3}-1}{x^2-3x+2}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\sqrt{2^2+3}-1}{2^2-3\cdot2+2}\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}\sqrt{2^2+3}-1=\sqrt{7}-1>0\\\lim\limits_{x\rightarrow2}2^2-3\cdot2+2=0\end{matrix}\right.\)

b: \(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{4x-1}+3}{x^2-4}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{4x-1-9}{\sqrt{4x-1}-3}\cdot\dfrac{1}{x^2-4}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{4x-10}{\sqrt{4x-1}-3}\cdot\dfrac{1}{\left(x-2\right)\left(x+2\right)}\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}\dfrac{4x-10}{\sqrt{4x-1}-3}=\dfrac{4\cdot2-10}{\sqrt{4\cdot2-1}-3}=\dfrac{-2}{\sqrt{7}-3}>0\\\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x-2\right)\cdot\left(x+2\right)}=\dfrac{1}{\left(2+2\right)\cdot\left(2-2\right)}=+\infty\end{matrix}\right.\)

I.

Do \(\left(u_n\right)\) là cấp số nhân \(\Rightarrow\)\(u_4=u_3.q\Rightarrow q=\dfrac{u_4}{u_3}=\dfrac{10}{3}\)

\(u_3=u_1q^2\Rightarrow u_1=\dfrac{u_3}{q^2}=\dfrac{27}{100}\)

2. Công thức số hạng tổng quát: \(u_n=\dfrac{27}{100}.\left(\dfrac{10}{3}\right)^{n-1}\)

II.

1. \(\lim\limits\dfrac{-3n^2+2n-2022}{3n^2-2022}=\lim\dfrac{-3+\dfrac{2}{n}-\dfrac{2022}{n^2}}{3-\dfrac{2022}{n^2}}=\dfrac{-3+0-0}{3-0}=-1\)

2.

\(\lim\limits_{x\rightarrow2}\dfrac{x^2-5x+6}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x-3\right)}{x-2}=\lim\limits_{x\rightarrow2}\left(x-3\right)=-1\)

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 = -∞.

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-\left(x+1\right)}{2x^2-x}=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{x^2+1}-\left(x+1\right)\right)\left(\sqrt{x^2+1}+x+1\right)}{x\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-2x}{x\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-2}{\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)

\(=\dfrac{-2}{\left(0-1\right)\left(\sqrt{1}+1\right)}=1\)

a. \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\lim\limits_{x\rightarrow2}\dfrac{1}{x+2}=\dfrac{1}{4}\)

b. \(\lim\limits_{x\rightarrow3^-}\dfrac{x+3}{x-3}=\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}\)

Do \(\lim\limits_{x\rightarrow3^-}\left(-x-3\right)=-6< 0\)

\(\lim\limits_{x\rightarrow3^-}\left(3-x\right)=0\) và \(3-x>0;\forall x< 3\)

\(\Rightarrow\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}=-\infty\)

a/ L'Hospital:

 \(=\lim\limits_{x\rightarrow2}\dfrac{x-\left(x+2\right)^{\dfrac{1}{2}}}{\left(4x+1\right)^{\dfrac{1}{2}}-3}=\lim\limits_{x\rightarrow2}\dfrac{1-\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}{\dfrac{1}{2}\left(4x+1\right)^{-\dfrac{1}{2}}.4}=\dfrac{1-\dfrac{1}{2}.4^{-\dfrac{1}{2}}}{2.9^{-\dfrac{1}{2}}}=\dfrac{9}{8}\)

b/ L'Hospital:\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7\right)^{\dfrac{1}{2}}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{2}\left(2x+7\right)^{-\dfrac{1}{2}}.2+1}{3x^2-8x}=\dfrac{9^{-\dfrac{1}{2}}+1}{3-8}=-\dfrac{4}{15}\)