\(\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} - x + 1} - x} \right) = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{{x^2} + x + 1 - {x^2}}}{{\sqrt {{x^2} + x + 1} + x}} = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{1 + \dfrac{1}{x}}}{{\sqrt {1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}} + 1} }} = \dfrac{1}{2}\)
a) lim ( x2+x-1)
x-> -∞
b) lim ( \(\sqrt{x^2+x+1}-2\sqrt{x^2-x}+x\))
x-> +∞
c) lim x\(\left(\sqrt{x^2+2x}-2\sqrt{x^2+x}+x\right)\)
x-> +∞
a) lim ( \(\sqrt{x^2-x+1}-\sqrt{x^2+x+1}\)
x-> +∞
b) lim \(\dfrac{\sqrt{4x+1}-3}{x^2-4}\)
x-> 2
c) lim \(\dfrac{\sqrt{2x+5}-1}{x^2-4}\)
x-> -2
a) lim \(\dfrac{x\sqrt{x^2+1}-2x+1}{^3\sqrt{2x^3-2}+1}\)
x-> -∞
b) lim \(\dfrac{\left(2x+1\right)^3\left(x+2\right)^4}{\left(3-2x\right)^7}\)
x-> -∞
c) lim \(\dfrac{\sqrt{4x^2+x}+^3\sqrt{8x^3+x-1}}{^4\sqrt{x^4+3}}\)
x-> +∞
a) lim (2x+ \(\sqrt{4x^2-x+1}\))
x-> -∞
b) lim x\(\left(\sqrt{4x^2+1}-x\right)\)
x-> -∞
a) lim \(\dfrac{2x-\sqrt{3x^2+2}}{5x+\sqrt{x^2+2}}\)
x-> +∞
b) lim \(\sqrt{\dfrac{x^2+1}{2x^4+x^2-3}}\)
x-> ∞
cho \(\lim\limits_{x\rightarrow-\infty}\dfrac{a\sqrt{x^2+1}+2017}{x+2018}=\dfrac{1}{2}\); \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+bx+1}-x\right)=2\). Tính P=4a+b
a) lim \(\dfrac{3x^4-2x^5}{5x^4+x+4}\)
x-> -∞
b) lim \(\dfrac{x-1}{\sqrt{x^2-1}}\)
x-> +∞
Tìm các giới hạn sau :
A=\(\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{x+1}-1}{\sqrt[4]{2x+1}-1}\)
B=\(\lim\limits_{x\rightarrow7}\frac{\sqrt[3]{4x-1}\sqrt{x-2}}{\sqrt[4]{2x+2}-2}\)
C=\(\lim\limits_{x\rightarrow0}\frac{\sqrt{\left(2x+1\right)\left(3x+1\right)\left(4x+1\right)}-1}{x}\)
D=\(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^2}\)
E=\(\lim\limits_{x\rightarrow0}\frac{\sqrt[m]{1+ax}-\sqrt[n]{1+bx}}{x}\)
Giup mình vớiii
Tìm giơi han:
a) lim (x-> \(+\infty\)) \(\dfrac{\sqrt{x^2+1}+x}{5-2x}\)
b) lim (x->4) \(\left(\dfrac{\sqrt{15x+4}-\sqrt{x-3}-3}{-x+4}\right)\)
sorry, e k bt nhâp lim ..
