- 1, [tex]\lim_{x\rightarrow 1}[/tex] [tex]\frac{\sqrt[3]{x+2}+x^{^{2}}+2x }{\sqrt{x}-1}[/tex]
- 2, [tex]\lim_{x\rightarrow 0}[/tex] [tex]\frac{\sqrt{1+2x}-\sqrt[3]{1+3x}}{x^{2}}[/tex]
- 3, [tex]\lim_{x\rightarrow 0}[/tex] [tex]\frac{\sqrt{1+x^{2}}-cos x}{x^{2}}[/tex]
- 4, [tex]\lim_{x\rightarrow 0}[/tex] [tex]\frac{\sqrt{5x^{3}+x^{2}+6x+9}-\sqrt[3]{9x^{2}+27x+27}}{x^{2}}[/tex]
CÁC BẠN ƠI GIÚP MK MẤY BÀI NÀY VS. CẢM ƠN NHIỀU 
r106
1. Bạn kiểm tra lại đề câu 1 nhé...
2.
$\lim_{x\rightarrow 0}\dfrac{\sqrt{1+2x}-\sqrt[3]{1+3x}}{x^{2}}=\lim_{x\rightarrow 0}\dfrac{\sqrt{1+2x}-(x+1)}{x^{2}}+\lim_{x\rightarrow 0}\dfrac{(x+1)-\sqrt[3]{1+3x}}{x^{2}}$
$=\lim_{x\rightarrow 0}\dfrac{-1}{\sqrt{1+2x}+x+1}+\lim_{x\rightarrow 0}\dfrac{x+3}{(x+1)^{2}+(x+1)\sqrt[3]{1+3x}+\sqrt[3]{(1+3x)^{2}}}=\dfrac{-1}{2}+1$
$=\dfrac{1}{2}$
3.
$\lim_{x\rightarrow 0}\dfrac{\sqrt{1+x^{2}}-cosx}{x^{2}}=\lim_{x\rightarrow 0}\dfrac{\sqrt{1+x^{2}}-1}{x^{2}}+\lim_{x\rightarrow 0}\dfrac{1-cosx}{x^{2}}$
$=\lim_{x\rightarrow 0}\dfrac{1}{\sqrt{1+x^{2}}+1}+\lim_{x\rightarrow 0}\dfrac{sin^{2}\frac{x}{2}}{\frac{x^{2}}{4}.2}=1$
4.
$\lim_{x\rightarrow 0}\dfrac{\sqrt{5x^{3}+x^{2}+6x+9}-\sqrt[3]{9x^{2}+27x+27}}{x^{2}}$
$=\lim_{x\rightarrow 0}\dfrac{\sqrt{5x^{3}+x^{2}+6x+9}-(x+3)}{x^{2}}+\lim_{x\rightarrow 0}\dfrac{(x+3)-\sqrt[3]{9x^{2}+27x+27}}{x^{2}}$
$=\lim_{x\rightarrow 0}\dfrac{5x}{\sqrt{5x^{3}+x^{2}+6x+9}+x+3}$
$+\lim_{x\rightarrow 0}\dfrac{x}{(x+3)^{2}+(x+3)\sqrt[3]{9x^{2}+27x+27}+\sqrt[3]{(9x^{2}+27x+27)^{2}}}=0$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
