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NV
27 tháng 3 2021

1.

\(\lim\dfrac{5\sqrt{3n^2+n}}{2\left(3n+2\right)}=\lim\dfrac{5\sqrt{3+\dfrac{1}{n}}}{2\left(3+\dfrac{2}{n}\right)}=\dfrac{5\sqrt{3}}{6}\Rightarrow a+b=11\)

2.

\(\lim\limits_{x\rightarrow2}\dfrac{x^2+ax+b}{x-2}=6\) khi \(x^2+ax+b=0\) có nghiệm \(x=2\)

\(\Rightarrow4+2a+b=0\Rightarrow b=-2a-4\)

\(\lim\limits_{x\rightarrow2}\dfrac{x^2+ax-2a-4}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)+a\left(x-2\right)}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+a+2\right)}{x-2}\)

\(=\lim\limits_{x\rightarrow2}\left(x+a+2\right)=a+4\Rightarrow a+4=6\Rightarrow a=2\Rightarrow b=-8\)

\(\Rightarrow a+b=-6\)

NV
2 tháng 3 2021

\(4x^3-3x+1=\left(2x-1\right)^2\left(x+1\right)\) có nghiệm kép \(x=\dfrac{1}{2}\)

\(\Rightarrow\sqrt{1+ax^2}-bx-2=0\) có nhiều hơn 1 nghiệm \(x=\dfrac{1}{2}\)

\(\Rightarrow\sqrt{1+\dfrac{a}{4}}=\dfrac{b}{2}+2\Rightarrow\sqrt{a+4}=b+4\) (\(b\ge-4\))

\(\Rightarrow a=b^2+8b+12\)

\(\Rightarrow\sqrt{1+\left(b^2+8b+12\right)x^2}=bx+2\)

\(\Rightarrow1+\left(b^2+8b+12\right)x^2=b^2x^2+4bx+4\)

\(\Rightarrow\left(8b+12\right)x^2-4bx-3=0\)

\(\Rightarrow\left(2x-1\right)\left[\left(4b+6\right)x+3\right]=0\)

\(\Rightarrow\left(4b+6\right)x+3=0\) có nghiệm \(x=\dfrac{1}{2}\)

\(\Rightarrow2b+3+3=0\Rightarrow b=-3\) \(\Rightarrow a=-3\)

Khi đó:

\(\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{\sqrt{1-3x^2}+3x-2}{4x^3-3x+1}=\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{-12\left(2x-1\right)^2}{\left(x+1\right)\left(2x-1\right)^2\left(\sqrt{1-3x^2}+2-3x\right)}\)

\(=\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{-12}{\left(x+1\right)\left(\sqrt{1-3x^2}+2-3x\right)}=-8\)

\(\Rightarrow c=-8\)

AH
Akai Haruma
Giáo viên
3 tháng 3 2021

Lời giải:

\(\lim\limits_{x\to 0,5}\frac{\sqrt{1+ax^2}-bx-2}{4x^3-3x+1}=\lim\limits_{x\to 0,5}\frac{\sqrt{1+ax^2}-bx-2}{(x+1)(2x-1)^2}\)

Để giới hạn hàm đã cho hữu hạn thì $f(x)=\sqrt{1+ax^2}-bx-2$ có nhân tử là $(2x-1)^2$

$f(x)$ có nhân tử $2x-1 \Leftrightarrow f(\frac{1}{2})=0\Leftrightarrow b=\sqrt{4+a}-4$

Khi đó:

$\sqrt{1+ax^2}-bx-2=(2x-1)(2-\frac{2x+1}{\sqrt{1+ax^2}+x\sqrt{4+a}})$

Giờ ta cần xác định $a,b$ để $2-\frac{2x+1}{\sqrt{1+ax^2}+x\sqrt{4+a}}=0$ với $x=\frac{1}{2}$

$\Leftrightarrow \sqrt{4+a}=1\Leftrightarrow a=-3$

$b=\sqrt{4+a}-4=-3$

\(\lim\limits_{x\to 0,5}\frac{\sqrt{1-3x^2}+3x-2}{4x^3-3x+1}=\lim\limits_{x\to 0,5}\frac{-3(2x-1)^2(2x+1)}{(2\sqrt{1-3x^2}+1)(\sqrt{1-3x^2}+x)(2x-1)^2(x+1)}\)

\(=\lim\limits_{x\to 0,5}\frac{-3(2x+1)}{(2\sqrt{1-3x^2}+1)(\sqrt{1-3x^2}+x)(x+1)}=-2=c\)

NV
27 tháng 1 2021

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

9 tháng 2 2021

a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x}{x}+\dfrac{3}{x}}{\dfrac{3x}{x}-\dfrac{1}{x}}=\dfrac{1}{3}\)

b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}+\dfrac{4}{x^2}}-\dfrac{x}{x}}{\dfrac{3x}{x}-\dfrac{1}{x}}=-\dfrac{2}{3}\)

a: \(=lim_{x->-\infty}\dfrac{2x-5+\dfrac{1}{x^2}}{7-\dfrac{1}{x}+\dfrac{4}{x^2}}\)

\(=\dfrac{2x-5}{7}\)

\(=\dfrac{2}{7}x-\dfrac{5}{7}\)

\(=-\infty\)

b: \(=lim_{x->+\infty}x\sqrt{\dfrac{1+\dfrac{1}{x}+\dfrac{3}{x^2}}{3x^2+4-\dfrac{5}{x^2}}}\)

\(=lim_{x->+\infty}x\sqrt{\dfrac{1}{3x^2+4}}=+\infty\)

9 tháng 2 2021

a/ \(\lim\limits_{x\rightarrow2}\dfrac{2+3}{4+2+4}=\dfrac{5}{10}=\dfrac{1}{2}\)

b/ \(\lim\limits_{x\rightarrow-3}\dfrac{\left(x+2\right)\left(x+3\right)}{x\left(x+3\right)}=\lim\limits_{x\rightarrow-3}\dfrac{x+2}{x}=\dfrac{-3+2}{-3}=\dfrac{1}{3}\)

AH
Akai Haruma
Giáo viên
14 tháng 5 2021

Lời giải:
a) 

\(\lim\limits_{x\to +\infty}\frac{\sqrt[3]{x^3+2x^2-4x+1}}{\sqrt{2x^2+x-8}}=\lim\limits_{x\to +\infty}\frac{\sqrt[3]{1+\frac{2}{x}-\frac{4}{x^2}+\frac{1}{x^3}}}{\sqrt{2+\frac{1}{x}-\frac{8}{x^2}}}\)

\(=\frac{1}{\sqrt{2}}\)

b) 

\(\lim\limits_{x\to -\infty}\frac{\sqrt{x^2-2x+4}-x}{3x-1}=\lim\limits_{x\to -\infty}\frac{\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}+1}{-3+\frac{1}{x}}=\frac{-1}{3}\)