tìm lim a, lim [{căn(4x^2 – x)} – 2x] x–>+vô cực b, lim (7x + x^2) / (x + 2) x–>-2- c, lim (x^4 + x^2) / (1- x^3) x–>1+

0 bình luận về “tìm lim a, lim [{căn(4x^2 – x)} – 2x] x–>+vô cực b, lim (7x + x^2) / (x + 2) x–>-2- c, lim (x^4 + x^2) / (1- x^3) x–>1+”

1. a) $\lim_{x  \to +\infty} \dfrac{4x^2-x-4x^2}{ \sqrt[]{4x^2-x}+2x } $

   =$\lim_{x \to+ \infty} \dfrac{-x} {\sqrt[]{4x^2-x}+2x} $

   = $\lim_{x \to+ \infty} \dfrac{-1}{ \sqrt[]{4- \dfrac{1}{x} }+2 } $

   =$\dfrac{-1}{2+2}= \dfrac{-1}{4}$

   b) $\lim_{x \to -2^-} \dfrac{7x+x^2}{x+2} $ 

   ta có $x<-2 <=>x+2<0$

   mà $\lim_{x \to-2^-} 7x+x^2 =7.-2+(-2)^2=-10<0$

   => $\lim_{x \to -2^-} \dfrac{7x+x^2}{x+2} =+∝$

   c) $\lim_{x\to 1^+} \dfrac{x^4+x^2}{1-x^3} $ 

   ta có $x>1 <=>x^3>1<=>-x^3<-1<=>1-x^3<0

   mà $\lim_{x \to1^+} x^4+x^2=1+1=2>0$

   => $\lim_{x \to 1^+} \dfrac{x^4+x^2}{1-x^3}=-∝ $

   hay thì xin hay nhất

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