Bài 2: Giới hạn của hàm số

Câu 1.

\(lim\left(\sqrt{n^2+2n+5}-\sqrt{n^2+n}\right)\)

Nhân liên hợp ta đc:

\(lim\left(\dfrac{n^2+2n+5-\left(n^2+n\right)}{\sqrt{n^2+2n+5}+\sqrt{n^2+n}}\right)\)

\(=lim\left(\dfrac{n+5}{n\sqrt{1+\dfrac{2}{n}+\dfrac{5}{n^2}}+n\sqrt{1+\dfrac{1}{n}}}\right)\)

\(=lim\left(\dfrac{n+\dfrac{5}{n}}{n\sqrt{1}+n\sqrt{1}}\right)=lim\left(\dfrac{n}{2n}\right)=lim\dfrac{1}{2}=\dfrac{1}{2}\)

Chọn C

\(\lim\limits_{x\rightarrow-3}\dfrac{x^4-6x^2-27}{x^3+3x^2+x+3}=\lim\limits_{x\rightarrow-3}\dfrac{\left(x+3\right)\left(x-3\right)\left(x^2+3\right)}{\left(x+3\right)\left(x^2+1\right)}=\lim\limits_{x\rightarrow-3}\dfrac{\left(x-3\right)\left(x^2+3\right)}{x^2+1}=\dfrac{-6.12}{10}=...\)

\(\lim\limits_{x\rightarrow0}\dfrac{4x}{\sqrt{9+x}-3}=\lim\limits_{x\rightarrow0}\dfrac{4x\left(\sqrt{9+x}+3\right)}{x}=\lim\limits_{x\rightarrow0}\dfrac{4\left(\sqrt{9+x}+3\right)}{1}=4\left(\sqrt{9}+3\right)=24\)

\(\lim\limits_{x\rightarrow1}\dfrac{3x-2\sqrt{4x^2-x-2}}{x^2-3x+2}=\dfrac{3-2\sqrt{4-1-2}}{0}=\dfrac{1}{0}=+\infty\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+7}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{2\left(x-1\right)}{\sqrt{2x+7}+3}+x-1}{\left(x-1\right)\left(x^2-3x-3\right)}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{2}{\sqrt{2x+7}+3}+1}{x^2-3x-3}=\dfrac{\dfrac{2}{\sqrt{9}+3}+1}{1-3-3}=-\dfrac{4}{15}\)

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+1}+\sqrt{x+4}-3}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x}{\sqrt{x+1}+1}+\dfrac{x}{\sqrt{x+4}+2}}{x}=\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sqrt{x+1}+1}+\dfrac{1}{\sqrt{x+4}+2}\right)=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4}\)

\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt[3]{4x}-2}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(\sqrt[3]{\left(4x\right)^2}+2\sqrt[3]{4x}+4\right)}=\lim\limits_{x\rightarrow2}\dfrac{4}{\sqrt[3]{\left(4x\right)^2}+2\sqrt[3]{4x}+4}=\dfrac{4}{\sqrt[3]{64}+2\sqrt[3]{8}+4}=...\)

Tham khảo:

Chúc bn học tốt

7.

\(\lim\left(3.4^n-5^n\right)=\lim5^n\left(3.\left(\dfrac{4}{5}\right)^n-1\right)=+\infty.\left(-1\right)=-\infty\)

8.

\(\lim\dfrac{n^2+n-1}{3n+2}=\lim\dfrac{n^2\left(1+\dfrac{1}{n}-\dfrac{1}{n^2}\right)}{n\left(3+\dfrac{2}{n}\right)}=\lim\dfrac{n\left(1+\dfrac{1}{n}-\dfrac{1}{n^2}\right)}{3+\dfrac{2}{n}}=\dfrac{+\infty}{3}=+\infty\)

1. \(lim_{n\rightarrow+\infty}\dfrac{n^2+1}{n}=lim_{n\rightarrow+\infty}\left(n+\dfrac{1}{n}\right)=+\infty\)(đpcm)

2. \(lim_{n\rightarrow+\infty}\dfrac{2-n}{\sqrt{n}}=lim_{n\rightarrow+\infty}\left(2-\sqrt{n}\right)=-\infty\) (đpcm)

a.