Đây ko phải dạng vô định, cứ thay số thôi bạn:
\(=\frac{1}{0}=+\infty\)
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Bài 2: Giới hạn của hàm số
Các câu hỏi tương tự
\(\lim\limits_{x\rightarrow1}\dfrac{x^3-3x^2+2}{x^2-4x+3}\)
\(\lim\limits_{x\rightarrow1^-}\dfrac{x^2+3x+2}{\left|x+1\right|}\)
\(\lim\limits_{x\rightarrow3}\dfrac{\sqrt[3]{x+5}-2}{x^2-4x+3}\)
a. limlimits_{xrightarrow0}frac{sqrt{1+2x}-1}{2x} f. limlimits_{xrightarrow1}frac{sqrt{2x+7-3}}{2-sqrt{x+3}}
b. limlimits_{xrightarrow0}frac{4x}{sqrt{9+x}-3} g. limlimits_{xrightarrow0}frac{sqrt{x^2+1}-1}{sqrt{x^2+16}-4}
c. limlimits_{xrightarrow2}frac{sqrt{x+7}-3}{x-2} h. limlimits_{xrightarrow4}frac{sqrt{x+5}-sqrt{2x+1}}{x-4}
d. limlimits_{xrightarrow1}frac{3x-2sqrt{4x^2-x-2}}{x^2-3x+2} k. limlimits...
Đọc tiếp
a. \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}-1}{2x}\) f. \(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7-3}}{2-\sqrt{x+3}}\)
b. \(\lim\limits_{x\rightarrow0}\frac{4x}{\sqrt{9+x}-3}\) g. \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}\)
c. \(\lim\limits_{x\rightarrow2}\frac{\sqrt{x+7}-3}{x-2}\) h. \(\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-\sqrt{2x+1}}{x-4}\)
d. \(\lim\limits_{x\rightarrow1}\frac{3x-2\sqrt{4x^2-x-2}}{x^2-3x+2}\) k. \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}+\sqrt{x+4}-3}{x}\)
e. \(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}+x-4}{x^3-4x^2+3}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-\sqrt[3]{2x+1}}{x}\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{4x+5}-3}{\sqrt[3]{5x+3}-2}\)
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt[4]{2x+3}+\sqrt[3]{2+3x}}{\sqrt{x+2}-1}\)
\(\lim\limits_{x\rightarrow-4}\dfrac{x^2+3x-4}{x^2+4x}\)
\(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x^2-x+3}}{2\left|x\right|-1}\)
Cho \(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-2x+1}{x-1}=3\)
Tính \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{3f\left(x\right)+1}-x-1}{\sqrt{4x+5}-3x-2}\)
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; \(\lim\limits_{x\rightarrow-1}\frac{\left(x+2\right)^2-1}{x^2-1}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{x^2+1}+2x+1}{\sqrt[3]{2x^3+x+1}+x}\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x^2-x+1}-\sqrt[3]{2x+3}}{3x^2-2}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{4x^2+x}+\sqrt[3]{8x^3+x-1}}{\sqrt[4]{x^4+3}}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\left(1+3x\right)^3-\left(1-4x\right)^4}{x}\)
\(\lim\limits_{x\rightarrow2}\dfrac{2x^2-5x+2}{x^3-3x-2}\)
\(\lim\limits_{x\rightarrow1}\dfrac{x^4-3x+2}{x^3+2x-3}\)
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)\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}-1}{x}\)