\(\lim\limits_{x\rightarrow0}\)\(\dfrac{2x}{\sqrt{2x^3+4x}}\)
\(=\lim\limits_{x\rightarrow0}\)\(\dfrac{2}{\sqrt{2x}+\dfrac{4}{x}}\)
`=0`
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8 tháng 1 lúc 20:56
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+2x}.\sqrt[3]{1+3x}-\sqrt{1+4x}}{1+x-\sqrt{1+2x}}=?\)
a) \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+2}+\sqrt{5x+4}-5}{x-1}_{ }\)
b) \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+4}+\sqrt{90-6x}-5}{x^2}\)
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) \(\lim\limits_{x\rightarrow3^-}\dfrac{x+3}{x-3}_{ }\)
29 tháng 1 2023 lúc 22:05
\(a,\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+1}-\sqrt{x^2+x+1}}{x^2-2x+1}\)
\(b,\lim\limits_{x\rightarrow7}\dfrac{\sqrt{x-3}-2}{49-x^2}\)
1) \(\lim\limits_{x\rightarrow4}\dfrac{\sqrt{2x+1}-\sqrt{x+5}}{x-4}\)
2) \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1-x}-\sqrt{1+x}}{x}\)
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) \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-\sqrt{x+2}}\)
tính giới hạn
a) \(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{2x+10}-4}{3x-9}\)
b) \(\lim\limits_{x\rightarrow7}\dfrac{\sqrt{4x+8}-6}{x^2-9x+14}\)
c) \(\lim\limits_{x\rightarrow5}\dfrac{x^2-8x+15}{2x^2-9x-5}\)
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) \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2-x+1}{x^3+x-2x^2}\)
tính giới hạn
a) \(\lim\limits_{x\rightarrow+\infty}\left(2x-\sqrt{x^2+4x-3}\right)\)
b) \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{4x^2-3x+1}-2x\right)\)