Tính các giới hạn sau :
a) \(\lim\limits_{x\rightarrow-3}\dfrac{x+3}{x^2+2x-3}\)
b) \(\lim\limits_{x\rightarrow0}\dfrac{\left(1+x\right)^3-1}{x}\)
c) \(\lim\limits_{x\rightarrow+\infty}\dfrac{x-1}{x^2-1}\)
d) \(\lim\limits_{x\rightarrow5}\dfrac{x-5}{\sqrt{x}-\sqrt{5}}\)
e) \(\lim\limits_{x\rightarrow+\infty}\dfrac{x-5}{\sqrt{x}+\sqrt{5}}\)
f) \(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x^2+5}-3}{x+2}\)
g) \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{x}-1}{\sqrt{x+3}-2}\)
h) \(\lim\limits_{x\rightarrow+\infty}\dfrac{1-2x+3x^3}{x^3-9}\)
i) \(\lim\limits_{x\rightarrow0}\dfrac{1}{x^2}\left(\dfrac{1}{x^2+1}-1\right)\)
j) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\left(x^2-1\right)\left(1-2x\right)^5}{x^7+x+3}\)
Tìm các giới hạn sau:
\(\lim\limits_{x\rightarrow-\infty}\) \(\dfrac{\sqrt{x^6+2}}{3\text{x}^3-1}\)
\(\lim\limits_{x\rightarrow+\infty}\) \(\dfrac{\sqrt{x^6+2}}{3\text{x}^3-1}\)
\(\lim\limits_{x\rightarrow-\infty}\) \(\left(\sqrt{2\text{x}^2+1}+x\right)\)
\(\lim\limits_{x\rightarrow1}\) \(\dfrac{2\text{x}^3-5\text{x}-4}{\left(x+1\right)^2}\)
Tính các giới hạn
a) \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+3}{3x-1}\)
b) \(\lim\limits_{x\rightarrow+\infty}\dfrac{\left(\sqrt{x^2+1}+x\right)^n-\left(\sqrt{x^2+1}-x\right)^n}{x}\)
Lời giải:
a) \(\lim\limits_{x\to -\infty}\frac{x+3}{3x-1}=\lim\limits_{x\to -\infty}\frac{1+\frac{3}{x}}{3-\frac{1}{x}}=\frac{1}{3}\)
b)
\(\lim\limits_{x\to +\infty}\frac{(\sqrt{x^2+1}+x)^n-(\sqrt{x^2+1}-x)^n}{x}=\lim\limits_{x\to +\infty} 2[(\sqrt{x^2+1}+x)^{n-1}+(\sqrt{x^2+1}+x)^{n-1}(\sqrt{x^2+1}-x)+....+(\sqrt{x^2+1}-x)^{n-1}]\)
\(=+\infty\)
Tính các giới hạn
a) \(\lim\limits_{x\rightarrow+\infty}\sqrt[3]{x^3+3x^2}-\sqrt{x^2-2x}\)
b) \(\lim\limits_{x\rightarrow+\infty}\sqrt[n]{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)}-x\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[n]{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)}-x\right)\\ =\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)-x^n}{\sqrt[n]{\left(\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)\right)^{n-1}}+...+x^{n-1}}\right)\)
= hệ số xn-1 trên tử/hệ số xn-1 dưới mẫu = \(\dfrac{a_1+a_2+...+a_n}{n}\)
Tìm các giới hạn sau :
a) \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{4-\sqrt{x^2+16}}\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\)
c) \(\lim\limits_{x\rightarrow+\infty}\dfrac{2x^4+5x-1}{1-x^2+x^4}\)
d) \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+\sqrt{4x^2-x+1}}{1-2x}\)
e) \(\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{x^2+1}-x\right)\)
f) \(\lim\limits_{x\rightarrow2^+}\left(\dfrac{1}{x^2-4}-\dfrac{1}{x-2}\right)\)
\(\lim\limits_{x\rightarrow0^-}\left(\dfrac{1}{x^2}-\dfrac{2}{x^3}\right)\)
\(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x^3-x^2}}{\sqrt{x-1}+1-x}\)
\(\lim\limits_{x\rightarrow1^+}\dfrac{1}{x^3-1}-\dfrac{1}{x-1}\)
\(\lim\limits_{x\rightarrow-\infty}\left(x-\sqrt[3]{1-x^3}\right)\)
1/ \(\lim\limits_{x\rightarrow0^-}\left(\dfrac{x-2}{x^3}\right)=\lim\limits_{x\rightarrow0^-}\dfrac{2-x}{-x^3}=\dfrac{2}{0}=+\infty\)
2/ \(\lim\limits_{x\rightarrow1^+}\dfrac{\left(x^3-x^2\right)^{\dfrac{1}{2}}}{\left(x-1\right)^{\dfrac{1}{2}}+1-x}=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{1}{2}\left(x^3-x^2\right)^{-\dfrac{1}{2}}.\left(3x^2-2x\right)}{\dfrac{1}{2}\left(x-1\right)^{-\dfrac{1}{2}}-1}=0\)
3/ \(\lim\limits_{x\rightarrow1^+}\dfrac{1-\left(x^2+x+1\right)}{x^3-1}=\dfrac{1-3}{0}=-\infty\)
4/ \(\lim\limits_{x\rightarrow-\infty}\left(-\infty-\sqrt[3]{1+\infty}\right)=-\left(\infty+\infty\right)=-\infty?\) Cái này ko chắc :v
Tính các giới hạn :
a) \(\lim\limits_{x\rightarrow+\infty}\left(\dfrac{x^3}{3x^2-4}-\dfrac{x^2}{3x+2}\right)\)
b) \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{9x^2+1}-3x\right)\)
c) \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{2x^2-3}-5x\right)\)
d) \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{2x^2+3}}{4x+2}\)e) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{2x^2+3}}{4x+2}\)
\(\lim\limits_{x\rightarrow-\infty}\left(x-\sqrt{x^2+x+1}\right)\)
\(\lim\limits_{x\rightarrow\pm\infty}\left(\sqrt{x^2+3x+1}-\sqrt{x^2-x+1}\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[3]{8x^3+2x}-2x\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[4]{16x^4+3x+1}-\sqrt{4x^2+2}\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}+\sqrt{x^2-x}-2x\right)\)
1/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2-x^2-x-x}{x+\sqrt{x^2+x+1}}=\dfrac{-2}{1-1}=-\infty\)
2/ tien toi +- vo cung?
3/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{8x^3+2x-8x^3}{\sqrt[3]{\left(8x^3+2x\right)^2}+2x.\sqrt[3]{8x^3+2x}+4x^2}=\dfrac{\dfrac{2x}{x^2}}{\dfrac{4x^2}{x^2}+\dfrac{4x^2}{x^2}+\dfrac{4x^2}{x^2}}=0\)
4/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{16x^4+3x+1-16x^4}{\sqrt[4]{\left(16x^4+3x+1\right)^3}+2x.\sqrt[4]{\left(16x^4+3x+1\right)^2}+4x^2.\sqrt[4]{16x^4+3x+1}+8x^3}+\lim\limits_{x\rightarrow+\infty}\dfrac{4x^2-4x^2-2}{2x+\sqrt{4x^2+2}}=\dfrac{\dfrac{3x}{x^3}}{8+8+8+8}-\dfrac{\dfrac{2}{x}}{2+2}=0\)
5/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}+\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-x-x^2}{\sqrt{x^2-x}+x}=\dfrac{\dfrac{1}{x}}{1+1}-\dfrac{\dfrac{x}{x}}{1+1}=-\dfrac{1}{2}\)
1, Tính:
a, \(\lim\limits_{x\rightarrow-2}\dfrac{x^3+2x^2}{\sqrt{x^2+4x+4}}\)
b, \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x+\sqrt{x+1}}-\sqrt{x}\right)\)
c, \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2-x}+1+\sqrt[3]{x^3+2}\right)\)
\(\lim\limits_{x\rightarrow-2}\dfrac{x^3+2x^2}{\sqrt{x^2+4x+4}}=\lim\limits_{x\rightarrow-2}\dfrac{x^2\left(x+2\right)}{\sqrt{\left(x+2\right)^2}}\)
\(=\lim\limits_{x\rightarrow-2}x^2=\left(-2\right)^2=4\)
p/s: bài này mình chưa học trên lớp nên ko chắc 100% đúng
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x+1}}{\sqrt{x+\sqrt{x+1}}+\sqrt{x}}=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x}}}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{1}{x^2}}}+1}=\dfrac{1}{1+1}=\dfrac{1}{2}\)
Câu c số 1 trong hay ngoài căn nhỉ?
