Ta có: \(0 \leqslant \left| {\cos 5x} \right| \leqslant 1 \Longrightarrow 0 \leqslant \left| {\dfrac{{\cos 5x}}{{2x}}} \right| \leqslant \dfrac{1}{{\left| {2x} \right|}},\forall x \ne 0\)
Mà: \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{1}{{\left| {2x} \right|}} = 0 \Longrightarrow \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\cos 5x}}{{2x}} = 0\)
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 \(\dfrac{2x-\sqrt{3x^2+2}}{5x+\sqrt{x^2+2}}\)
x-> +∞
b) lim \(\sqrt{\dfrac{x^2+1}{2x^4+x^2-3}}\)
x-> ∞
a) lim \(\dfrac{3x^4-2x^5}{5x^4+x+4}\)
x-> -∞
b) lim \(\dfrac{x-1}{\sqrt{x^2-1}}\)
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
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 ..
\(\lim\limits_{x\rightarrow1}\dfrac{3-2x^2-x^4}{5-3x-2x^2}\)
lim \(\dfrac{3x-5sin2x+cos^2x}{x^2+2}\)
x-> +∞
lim \(\dfrac{5x^2+2x+3}{x^2+1}\)
x-> +∞
lim \(\dfrac{-2x^5+x^4-3}{3x^2-7}\)
x-> -∞
