Bài 2: Giới hạn của hàm số

Ta có : \(A=lim_{x\rightarrow1}\dfrac{\sqrt{2x-1}-x}{x^2-1}=lim_{x\rightarrow1}\dfrac{2x-1-x^2}{\left(x^2-1\right)\left(\sqrt{2x-1}+x\right)}\)  \(=lim_{x\rightarrow1}\dfrac{-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)\left(\sqrt{2x-1}+x\right)}\)  \(=lim_{x\rightarrow1}\dfrac{1-x}{\left(x+1\right)\left(\sqrt{2x-1}+x\right)}=0\)

Đặt \(f\left(x\right)=\sqrt{x^2-ax}-b\)  \(\Rightarrow f\left(1\right)=\sqrt{1-a}-b\)

Ta có : \(Lim_{x\rightarrow1}\dfrac{f\left(x\right)}{x-1}=\dfrac{5}{4}\Rightarrow f\left(1\right)=0\)  \(\Leftrightarrow\sqrt{1-a}-b=0\Leftrightarrow\sqrt{1-a}=b\)

\(\Leftrightarrow1-a=b^2\Leftrightarrow a=1-b^2\) 

Khi đó : \(f\left(x\right)=\sqrt{x^2-\left(1-b^2\right)x}-b\)  \(=\dfrac{x^2-\left(1-b^2\right)x-b^2}{\sqrt{x^2-\left(1-b^2\right)x}+b}=\dfrac{x\left(x-1\right)+b^2\left(x-1\right)}{\sqrt{x^2-\left(1-b^2\right)x}+b}\) 

\(=\dfrac{\left(x+b^2\right)\left(x-1\right)}{\sqrt{x^2-\left(1-b^2\right)x}+b}\)

\(Lim_{x\rightarrow1}\dfrac{f\left(x\right)}{x-1}=Lim_{x\rightarrow1}\dfrac{x+b^2}{\sqrt{x^2-\left(1-b^2\right)x}+b}=\dfrac{1+b^2}{\sqrt{1-\left(1-b^2\right)}+b}=\dfrac{1+b^2}{b+\left|b\right|}=\dfrac{5}{4}\)\(\Rightarrow4\left(1+b^2\right)=5b+5\left|b\right|\)  

b < 0 -> Loại 

b \(\ge0\) 0 : \(4+4b^2=10b\Leftrightarrow2b^2-5b+2=0\) \(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=\dfrac{1}{2}\end{matrix}\right.\) 

b = 2 \(\Rightarrow a=1-2^2=-3\)

b = 1/2 \(\Rightarrow a=1-\left(\dfrac{1}{2}\right)^2=\dfrac{3}{4}\)

Vậy ... 

\(\lim\limits_{x\rightarrow1}\dfrac{x-1}{\sqrt{2-x}-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(\sqrt{2-x}+1\right)}{1-x}\)

\(=\lim\limits_{x\rightarrow1}\left(-\sqrt{2-x}-1\right)=-2\)

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4-x-x^2}-2}{x^2-x}=\lim\limits_{x\rightarrow0}\dfrac{-x^2-x}{x\left(x-1\right)\left(\sqrt{4-x-x^2}+2\right)}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-x-1}{\left(x-1\right)\left(\sqrt{4-x-x^2}+2\right)}=\dfrac{-1}{-1\left(2+2\right)}=\dfrac{1}{4}\)

Tính \(\lim_{x\to -\infty} ((2x+1)^2+4\sqrt{x^2+4}\sqrt[3]{x^3+3x^2})\)

ms đúng chứ

8a.

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\left(3x^2-5x+1\right)=3-5+1=-1\)

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(-3x+2\right)=-3+2=-1\)

\(\Rightarrow\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)\Rightarrow\) hàm có giới hạn tại \(x=1\)

Đồng thời \(\lim\limits_{x\rightarrow1}f\left(x\right)=-1\)

b.

\(\lim\limits_{x\rightarrow2^+}f\left(x\right)=\lim\limits_{x\rightarrow2^+}\dfrac{x^3-8}{x-2}=\lim\limits_{x\rightarrow2^+}\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{x-2}\)

\(=\lim\limits_{x\rightarrow2^+}\left(x^2+2x+4\right)=12\)

\(\lim\limits_{x\rightarrow2^-}f\left(x\right)=\lim\limits_{x\rightarrow2^-}\left(2x+1\right)=5\)

\(\Rightarrow\lim\limits_{x\rightarrow2^+}f\left(x\right)\ne\lim\limits_{x\rightarrow2^-}f\left(x\right)\Rightarrow\) hàm ko có giới hạn tại x=2

9.

\(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^+}\dfrac{x^2+mx+2m+1}{x+1}=\dfrac{0+0+2m+1}{0+1}=2m+1\)

\(\lim\limits_{x\rightarrow0^-}f\left(x\right)=\lim\limits_{x\rightarrow0^-}\dfrac{2x+3m-1}{\sqrt{1-x}+2}=\dfrac{0+3m-1}{1+2}=\dfrac{3m-1}{3}\)

Hàm có giới hạn khi \(x\rightarrow0\) khi:

\(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^-}f\left(x\right)\Rightarrow2m+1=\dfrac{3m-1}{3}\)

\(\Rightarrow m=-\dfrac{4}{3}\)

\(\lim\limits_{x\rightarrow4}\dfrac{\sqrt{x+5}-3}{x-4}=\lim\limits_{x\rightarrow4}\dfrac{\left(\sqrt{x+5}-3\right)\left(\sqrt{x+5}+3\right)}{\left(x-4\right)\left(\sqrt{x+5}+3\right)}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{x-4}{\left(x-4\right)\left(\sqrt{x+5}+3\right)}=\lim\limits_{x\rightarrow4}\dfrac{1}{\sqrt{x+5}+3}=\dfrac{1}{3+3}=\dfrac{1}{6}\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{x^2+3x}-2}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt{x^2+3x}-2\right)\left(\sqrt{x^2+3x}+2\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{x^2+3x}+2\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{x^2+3x-4}{\left(x-1\right)\left(x+1\right)\left(\sqrt{x^2+3x}+2\right)}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+4\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{x^2+3x}+2\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{x+4}{\left(x+1\right)\left(\sqrt{x^2+3x}+2\right)}=\dfrac{5}{2\left(2+2\right)}=\dfrac{5}{8}\)

\(\lim\limits_{x\rightarrow2}\dfrac{4x+3}{x-1}=\dfrac{4.2+3}{2-1}=11\)