Chọn D
lim x → 1 x 3 + 3 x + 12 = 1 3 + 3.1 + 12 = 4
Chọn D
lim x → 1 x 3 + 3 x + 12 = 1 3 + 3.1 + 12 = 4
11) \(\lim\limits_{x->1}\) \(\dfrac{3_{\sqrt{4x-1}-\sqrt{4x-3}}}{x-1}\)
11) \(\lim\limits_{x->4}\dfrac{4x-1}{x^2-8x+16}\)
12) \(\lim\limits_{x->2}\)\(\dfrac{4-x^2}{x^3-8}\)
13) \(\lim\limits_{x->+\infty}\left(3_{\sqrt{x^3+4x^2}-x}\right)\)
a) lim\(\dfrac{x^2-1}{x+1}\)(x-->-3)
b) lim\(\dfrac{4-x^2}{x+2}\)(x-->-2)
tính giới hạn
a) \(\lim\limits_{x\rightarrow4}\dfrac{\sqrt{2x+8}-4}{x-4}\)
b) \(\lim\limits_{x\rightarrow2}\dfrac{x^2-4}{\sqrt{4x+1}-3}\)
c) \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-\sqrt{x+2}}\)
4. Tính giới hạn \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-x-1}{2x^2-x}_{ }\)
5. Tính giới hạn:
a) \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{x^2-4}_{ }\)
b) \(\lim\limits_{x\rightarrow3^-}\dfrac{x+3}{x-3}_{ }\)
tính giới hạn
a) \(\lim\limits_{x\rightarrow+\infty}\dfrac{5x^2+x^3+5}{4x^3+1}\)
b) \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2-x+1}{x^3+x-2x^2}\)
c) \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2-x+1}{x^3+x-2x^2}\)
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 giới hạn
a) \(\lim\limits_{x\rightarrow-2}\dfrac{4-x^2}{2x^2+7x+6}\)
b) \(\lim\limits_{x\rightarrow4}\dfrac{2x^2-13x+20}{x^3+64}\)
c) \(\lim\limits_{x\rightarrow-1}\dfrac{2x^2+8x+6}{-2x^2+7x+9}\)
Tìm các số thực a, b thoả mãn:
\(\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left[\left(a^3+b^3\right)x^2-\left(x+a^2b\right)\sqrt{x^2+2\left(ab\right)^2}\right]}{x-b-1}\)
Giúp mình với ạ
1) lim\(\dfrac{3x^2+5}{x^3-x+2}\)(x-->+∞)
2) lim\(\dfrac{2x^2\left(3x^2-5\right)^3\left(1-x\right)^5}{3x^{14}+x^2-1}\)(x-->-∞)
3) lim\(\dfrac{3x-\sqrt{2x^2+5}}{x^2-4}\)(x-->+∞)
\(a,\lim\limits_{x\rightarrow2}\dfrac{x^3+2x^2-6x-4}{8-x^3}\)
\(b,\lim\limits_{x\rightarrow2}\dfrac{x^3+x^2-5x-2}{x^2-3x+2}\)