1 giới hạn bt ko phải dạng vô định, cứ thay số thôi bạn:
\(=\frac{-2-3+1}{0}=\frac{-4}{0}=-\infty\)
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Bài 2: Giới hạn của hàm số
Các câu hỏi tương tự
tìm các giới hạn sau:
a; \(\lim\limits_{x\rightarrow1}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\)
b; \(\lim\limits_{x\rightarrow1}\frac{x^3-3x^2+2}{x^4-4x+3}\)
c; \(\lim\limits_{x\rightarrow1}\frac{x^3-2x-1}{x^5-2x-1}\)
d; \(\lim\limits_{x\rightarrow-1}\frac{\left(x+2\right)^2-1}{x^2-1}\)
Bài 1
a. limlimits_{xrightarrow-infty}frac{sqrt{4x^2}+1}{3x-1}
b. limlimits_{xrightarrow+infty}frac{sqrt{9x^2+x+1}-sqrt{4x^2+2x+1}}{x+1}
c. limlimits_{xrightarrow+infty}frac{sqrt{x+2x+3}+4x+1}{sqrt{4x^2+1}+2-x}
d. limlimits_{xrightarrow+infty}frac{3x-2sqrt{x}+sqrt{x^4-5x}}{2x^2+4x-5}
Bài 2
a. limlimits_{xrightarrow-infty}frac{2x+1}{x-1}
b. limlimits_{xrightarrow-infty}frac{2x^3+3}{x^3-2x^2+1}
c. limlimits_{xrightarrow+infty}frac{left(3x^2+1right)left(5x+3right)}{left(2x^3-1r...
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Bài 1
a. \(\lim\limits_{x\rightarrow-\infty}\frac{\sqrt{4x^2}+1}{3x-1}\)
b. \(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9x^2+x+1}-\sqrt{4x^2+2x+1}}{x+1}\)
c. \(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+2x+3}+4x+1}{\sqrt{4x^2+1}+2-x}\)
d. \(\lim\limits_{x\rightarrow+\infty}\frac{3x-2\sqrt{x}+\sqrt{x^4-5x}}{2x^2+4x-5}\)
Bài 2
a. \(\lim\limits_{x\rightarrow-\infty}\frac{2x+1}{x-1}\)
b. \(\lim\limits_{x\rightarrow-\infty}\frac{2x^3+3}{x^3-2x^2+1}\)
c. \(\lim\limits_{x\rightarrow+\infty}\frac{\left(3x^2+1\right)\left(5x+3\right)}{\left(2x^3-1\right)\left(x+4\right)}\)
tìm các giới hạn sau:
a; \(\lim\limits_{x\rightarrow\frac{\pi}{2}}\frac{sin\left(x-\frac{\pi}{4}\right)}{x}\)
b, \(\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{3x^2-4}-\sqrt{3x-2}}{x+1}\)
c,\(\lim\limits_{x\rightarrow0}x^2sin\frac{1}{2}\)
1, limlimits_{xrightarrow1}frac{2x^2-3x+1}{x^3-x^2-x+1}
2, limlimits_{xrightarrow2}frac{x-sqrt{x+2}}{sqrt{4x+1}-3}
3, limlimits_{xrightarrow0}frac{1-sqrt[3]{x-1}}{x}
4, limlimits_{xrightarrow-infty}frac{x^2-5x+1}{x^2-2}
5, limlimits_{xrightarrow+infty}frac{2x^2-4}{x^3+3x^2-9}
6, limlimits_{xrightarrow2^-}frac{2x-1}{x-2}
7, limlimits_{xrightarrow3^+}frac{8+x-x^2}{x-3}
8, limlimits_{xrightarrow-infty}left(8+4x-x^3right)
9, limlimits_{xrightarrow-1}frac{sq...
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1, \(\lim\limits_{x\rightarrow1}\frac{2x^2-3x+1}{x^3-x^2-x+1}\)
2, \(\lim\limits_{x\rightarrow2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}\)
3, \(\lim\limits_{x\rightarrow0}\frac{1-\sqrt[3]{x-1}}{x}\)
4, \(\lim\limits_{x\rightarrow-\infty}\frac{x^2-5x+1}{x^2-2}\)
5, \(\lim\limits_{x\rightarrow+\infty}\frac{2x^2-4}{x^3+3x^2-9}\)
6, \(\lim\limits_{x\rightarrow2^-}\frac{2x-1}{x-2}\)
7, \(\lim\limits_{x\rightarrow3^+}\frac{8+x-x^2}{x-3}\)
8, \(\lim\limits_{x\rightarrow-\infty}\left(8+4x-x^3\right)\)
9, \(\lim\limits_{x\rightarrow-1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}\)
10, \(\lim\limits_{x\rightarrow-\infty}\frac{\left(2x^2+1\right)^2\left(5x+3\right)}{\left(2x^3-1\right)\left(x+1\right)^2}\)
11, \(\lim\limits_{x\rightarrow-\infty}\frac{\sqrt{x^2+2x}}{x+3}\)
12, \(\lim\limits_{x\rightarrow1}\frac{\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}\)
13, \(\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{x+1}+\sqrt{x+4}-3}{x}\)
14, \(\lim\limits_{x\rightarrow0}\frac{\left(x^2+2020\right)\sqrt{1+3x}-2020}{x}\)
15, \(\lim\limits_{x\rightarrow+\infty}\left(2x-\sqrt{4x^2-3}\right)\)
16, \(\lim\limits_{x\rightarrow a}\frac{x^2-\left(a+1\right)x+a}{x^3-a^3}\)
17, \(\lim\limits_{x\rightarrow1}\frac{x^n-nx+n-1}{\left(x-1\right)^2}\)
18, \(f\left(x\right)=\left\{{}\begin{matrix}\frac{x^2-2x}{8-x^3}\\\frac{x^4-16}{x-2}\end{matrix}\right.\) khi x>2,khi x<2 tại x=2
Tính các giới hạn sau :
a) limlimits_{xrightarrow-3}dfrac{x+3}{x^2+2x-3}
b) limlimits_{xrightarrow0}dfrac{left(1+xright)^3-1}{x}
c) limlimits_{xrightarrow+infty}dfrac{x-1}{x^2-1}
d) limlimits_{xrightarrow5}dfrac{x-5}{sqrt{x}-sqrt{5}}
e) limlimits_{xrightarrow+infty}dfrac{x-5}{sqrt{x}+sqrt{5}}
f) limlimits_{xrightarrow-2}dfrac{sqrt{x^2+5}-3}{x+2}
g) limlimits_{xrightarrow1}dfrac{sqrt{x}-1}{sqrt{x+3}-2}
h) limlimits_{xrightarrow+infty}dfrac{1-2x+3x^3}{x^3-9}
i) limlimits_{xrightarrow0}dfr...
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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) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\left(x^2-1\right)\left(1-2x\right)^5}{x^7+x+3}\)
Tính các giới hạn sau:
1. \(\lim\limits_{x\rightarrow a}\dfrac{x^2-\left(a+1\right)x+a}{x^3-a^3}\)
2. \(\lim\limits_{x\rightarrow1}\left(\dfrac{1}{1-x}-\dfrac{3}{1-x^3}\right)\)
3. \(\lim\limits_{h\rightarrow0}\dfrac{\left(x+h\right)^3-x^3}{h}\)
\(\lim\limits_{x\rightarrow-\infty}\left(3x^3+5x^2-9\sqrt{2}x-2017\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+x+1}-\sqrt[3]{2x^3+x-1}\right)\)
\(\lim\limits_{x\rightarrow-\infty}\left(x-\sqrt{x^2+x+1}\right)\)
\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+x^2+1}+\sqrt{x^2+x+1}\right)\)
Tìm các giới hạn sau:
C=\(\lim\limits_{x\rightarrow0}\frac{\left(3x+1\right)^3-\left(1-4x\right)^4}{x}\)
D=\(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)\left(1+2x\right)\left(1+3x\right)-1}{x}\)
tìm các giới hạn sau:
a; limlimits_{xrightarrow2}frac{sqrt[3]{8x+11}-sqrt{x+7}}{x^2-3x+2}
b, limlimits_{xrightarrow+infty}frac{left(2x-3right)^2left(4x+7right)^3}{left(3x^3+1right)left(10x^2+9right)}
c,limlimits_{xrightarrow0}frac{sqrt{1+4x}-sqrt[3]{1+6x}}{x^2} ( bài này k hiểu mk tính kiểu gì 1 cái ra +infty một cái ra -infty)
d, limlimits_{xrightarrow0}frac{sqrt{1+4x}.sqrt{1+6x}-1}{x}
e, limlimits_{xrightarrow0}frac{sqrt{1+2x}.sqrt[3]{1+4x}-1}{x}
Đọc tiếp
tìm các giới hạn sau:
a; \(\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^2-3x+2}\)
b, \(\lim\limits_{x\rightarrow+\infty}\frac{\left(2x-3\right)^2\left(4x+7\right)^3}{\left(3x^3+1\right)\left(10x^2+9\right)}\)
c,\(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^2}\) ( bài này k hiểu mk tính kiểu gì 1 cái ra \(+\infty\) một cái ra \(-\infty\))
d, \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+4x}.\sqrt{1+6x}-1}{x}\)
e, \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}.\sqrt[3]{1+4x}-1}{x}\)