Bài 2: Giới hạn của hàm số
Các câu hỏi tương tự
tính giới hạn sau:
\(\lim\limits_{x\rightarrow\dfrac{\pi}{6}}\dfrac{\sqrt{3}sinx-cosx}{sin\left(\dfrac{\pi}{3}-2x\right)}\)
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...
Đọc tiếp
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) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\left(x^2-1\right)\left(1-2x\right)^5}{x^7+x+3}\)
7 tháng 11 2023 lúc 19:44
limlimits_{xrightarrow2}left(dfrac{1}{x-2}-dfrac{12}{x^3-8}right)limlimits_{xrightarrow2}left(dfrac{1}{x^2-3x+2}+dfrac{1}{x^2-5x+6}right)limlimits_{xrightarrow+infty}xleft(sqrt{x^2+1}-xright)limlimits_{xrightarrow+infty}left(sqrt{x^2+1}-sqrt[3]{x^3-1}right)limlimits_{xrightarrow0}dfrac{sqrt{x^2+1}-1}{sqrt{x^2+16}-4}
Đọc tiếp
\(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x-2}-\dfrac{12}{x^3-8}\right)\)
\(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\)
\(\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{x^2+1}-x\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-\sqrt[3]{x^3-1}\right)\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}\)
23 tháng 1 2021 lúc 20:29
Cho f(x) thỏa mãn : \(_{\lim\limits_{x\rightarrow-1}\dfrac{2f\left(x\right)+1}{x+1}=5}\)
Tính I= \(\lim\limits_{x\rightarrow-1}\dfrac{\left(4f\left(x\right)+3\right)\left(\sqrt{4f\left(x\right)^2+2f\left(x\right)+4}\right)-2}{x^2-1}\)
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, \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2-x}+1+\sqrt[3]{x^3+2}\right)\)
\(\lim\limits_{x\rightarrow-4}\dfrac{x^2+3x-4}{x^2+4x}\)
\(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x^2-x+3}}{2\left|x\right|-1}\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2-2}+\sqrt[3]{x^3+1}}{\sqrt{x^2+1}-x}\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{2x+3}{\sqrt{2x^2-3}}\)
\(\lim\limits_{x\rightarrow\pm\infty}\dfrac{2x^2-1}{3-x^2}\)
\(\lim\limits_{x\rightarrow+\infty}\left(x+1\right)\left(\sqrt{\dfrac{x}{2x^4+x^2+1}}\right)\)
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. \(\lim\limits_{h\rightarrow0}\dfrac{\left(x+h\right)^3-x^3}{h}\)