Đây không phải giới hạn dạng vô định mà chỉ là giới hạn bình thường
\(=\frac{\sqrt[3]{19}-2\sqrt{2}}{0}=-\infty\)
Bài 2: Giới hạn của hàm số
Các câu hỏi tương tự
\(lim_{x->\frac{+}{ }\infty}\frac{\sqrt{x^2+3x+5}}{\sqrt[3]{x^3+7x^2+8}}\)
\(\lim_{x\to -\infty} ((2x+1)^2+4\sqrt{x^2+4}\sqrt[3]{x^3+3x^2})\)
a,^{lim}_{x-2}frac{sqrt[3]{8x+11}-sqrt{x+7}}{x^2-3x+2}
b, ^{lim}_{x-0}frac{2sqrt{1+x}-sqrt[3]{8-x}}{x}
c, ^{lim}_{x-1}frac{sqrt{5-x^3}-sqrt[3]{x^2+7}}{x^2-1}
d,^{lim}_{x-0}frac{sqrt{1+2x}.sqrt[3]{1+4x}-1}{x}
e,^{lim}_{x-1}frac{x^4-1}{x^3-2x^2+x}
f,^{lim}_{x-1}left(frac{1}{1-x}-frac{3}{1-x^3}right)
Đọc tiếp
a,\(^{lim}_{x->2}\frac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^2-3x+2}\)
b, \(^{lim}_{x->0}\frac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)
c, \(^{lim}_{x->1}\frac{\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}\)
d,\(^{lim}_{x->0}\frac{\sqrt{1+2x}.\sqrt[3]{1+4x}-1}{x}\)
e,\(^{lim}_{x->1}\frac{x^4-1}{x^3-2x^2+x}\)
f,\(^{lim}_{x->1}\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right)\)
\(Lim_{x\to3}\)\(\frac{2 - \sqrt(x+1)\sqrt[3](x-2)}{2- \sqrt(x-2)\sqrt[3](x+5)}\)
\(lim_{x\rightarrow\left(-1\right)^+}\left(x^3+1\right)\left(\sqrt{\dfrac{3x}{x^2-1}}\right)\)
a. limlimits_{xrightarrow a}frac{xsqrt{x}-asqrt{a}}{sqrt{x}-sqrt{a}} e. limlimits_{xrightarrow0}frac{sqrt{1+x}-sqrt[3]{1+x}}{x}
b. limlimits_{xrightarrow1}frac{sqrt[n]{x}-1}{sqrt[m]{x}-1}left(m,nin Z^+right) f. limlimits_{xrightarrow2}frac{sqrt[3]{8x+11}-sqrt{x+7}}{x^2-3x+2}
c. limlimits_{xrightarrow1}frac{left(1-sqrt{x}right)left(1-sqrt[3]{x}right)left(1-sqrt[4]{x}right)left(1-sqrt[5]{x}right)}{left(1-xright)^4}...
Đọc tiếp
a. \(\lim\limits_{x\rightarrow a}\frac{x\sqrt{x}-a\sqrt{a}}{\sqrt{x}-\sqrt{a}}\) e. \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}\)
b. \(\lim\limits_{x\rightarrow1}\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}\left(m,n\in Z^+\right)\) f. \(\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^2-3x+2}\)
c. \(\lim\limits_{x\rightarrow1}\frac{\left(1-\sqrt{x}\right)\left(1-\sqrt[3]{x}\right)\left(1-\sqrt[4]{x}\right)\left(1-\sqrt[5]{x}\right)}{\left(1-x\right)^4}\) g. \(\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{3x-2}-\sqrt{2x-1}}{x^3-1}\)
d. \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)\) h. \(\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+9}+\sqrt[3]{2x-6}}{x^3+1}\)
Tìm \(\lim_{x\to 1}\frac{\sqrt[7]{2-x}-1}{x-1}\)
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, \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}.\sqrt[3]{1+4x}-1}{x}\)
1) limlimits_{xrightarrow0}dfrac{2sqrt{1+x}-sqrt[3]{8-x}}{x}
2)limlimits_{xrightarrow1}dfrac{sqrt[3]{x+7}-sqrt{x+3}}{x^2-3x+2}
3)limlimits_{xrightarrow1}dfrac{sqrt[3]{x^2+7}-sqrt{5-x^2}}{x^2-1}
4)limlimits_{xrightarrow-2}dfrac{sqrt{x+11}-sqrt[3]{8x+43}}{2x^2+3x-2}
5) limlimits_{xrightarrow0}dfrac{sqrt[n]{1+ax}-sqrt[m]{1+bx}}{x}
6)limlimits_{xrightarrow0}dfrac{sqrt{1+4x}.sqrt[3]{1+6x}-1}{x}
Đọc tiếp
1) \(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)
2)\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-\sqrt{x+3}}{x^2-3x+2}\)
3)\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-\sqrt{5-x^2}}{x^2-1}\)
4)\(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x+11}-\sqrt[3]{8x+43}}{2x^2+3x-2}\)
5) \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+ax}-\sqrt[m]{1+bx}}{x}\)
6)\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}-1}{x}\)