a, lim 3x^2+1/2-x x->2- b,lim (√1+x-√x) x-> dương vô cùng 19/09/2021 Bởi Maya a, lim 3x^2+1/2-x x->2- b,lim (√1+x-√x) x-> dương vô cùng
`a, lim_{x->2^-} (3x^2+1)/(2-x)` `= lim_{x->2^-}[(3x^2+1)(2-x)^{-1}]` `= lim_{x->2^-} (3x^2 + 1) . lim_{x->2^-} [(2-x)^{-1}]` `= 13 . oo` `= oo` `b, lim_{x->+oo} \sqrt{1+x} – \sqrt{x}` `= lim_{x->+oo} 1/(\sqrt{1+x}+\sqrt{x})` `= (lim_{x->+oo} (1))/(lim_{x->+oo} (\sqrt{1+x}+\sqrt{x}))` `= 1/oo` `= 0` Bình luận
Đáp án: $a)\lim_{x \to 2^-}\dfrac{3x^2+1}{2-x}$Ta có: $\lim_{x \to 2^-} (2-x)=0; 2-x<0$ $\lim_{x \to 2^-} (3x^2+1)=3.2^2+1=13>0$ Vậy $\lim_{x \to 2^-}\dfrac{3x^2+1}{2-x}=-\infty$ $b)\lim_{x \to +\infty} $ $\sqrt[]{1+x}-\sqrt[]{x}$ $=\lim_{x \to +\infty}\dfrac{(\sqrt[]{1+x}-\sqrt[]{x})(\sqrt[]{1+x}+\sqrt[]{x})}{\sqrt[]{1+x}+\sqrt[]{x}} $ $=\lim_{x \to +\infty}\dfrac{1+x-x}{\sqrt[]{1+x}+\sqrt[]{x}} $ $=\lim_{x \to +\infty} \dfrac{1}{\sqrt[]{1+x}+\sqrt[]{x}}=0$ BẠN THAM KHẢO. Bình luận
`a, lim_{x->2^-} (3x^2+1)/(2-x)`
`= lim_{x->2^-}[(3x^2+1)(2-x)^{-1}]`
`= lim_{x->2^-} (3x^2 + 1) . lim_{x->2^-} [(2-x)^{-1}]`
`= 13 . oo`
`= oo`
`b, lim_{x->+oo} \sqrt{1+x} – \sqrt{x}`
`= lim_{x->+oo} 1/(\sqrt{1+x}+\sqrt{x})`
`= (lim_{x->+oo} (1))/(lim_{x->+oo} (\sqrt{1+x}+\sqrt{x}))`
`= 1/oo`
`= 0`
Đáp án:
$a)\lim_{x \to 2^-}\dfrac{3x^2+1}{2-x}$
Ta có:
$\lim_{x \to 2^-} (2-x)=0; 2-x<0$
$\lim_{x \to 2^-} (3x^2+1)=3.2^2+1=13>0$
Vậy $\lim_{x \to 2^-}\dfrac{3x^2+1}{2-x}=-\infty$
$b)\lim_{x \to +\infty} $ $\sqrt[]{1+x}-\sqrt[]{x}$
$=\lim_{x \to +\infty}\dfrac{(\sqrt[]{1+x}-\sqrt[]{x})(\sqrt[]{1+x}+\sqrt[]{x})}{\sqrt[]{1+x}+\sqrt[]{x}} $
$=\lim_{x \to +\infty}\dfrac{1+x-x}{\sqrt[]{1+x}+\sqrt[]{x}} $
$=\lim_{x \to +\infty} \dfrac{1}{\sqrt[]{1+x}+\sqrt[]{x}}=0$
BẠN THAM KHẢO.