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\(\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\left(a+1\right)x^3+bx^2-2ax-2b+1}{x^2-2}\right)\)
Giới hạn hữu hạn khi \(a+1=0\Rightarrow a=-1\)
\(\lim\limits_{x\rightarrow+\infty}\left(\dfrac{bx^2+2x-2b+1}{x^2-2}\right)=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{b+\dfrac{2}{x}-\dfrac{2b-1}{x^2}}{1-\dfrac{2}{x^2}}\right)=b\)
\(\Rightarrow b=10\)
Lời giải:\(\lim\limits_{x\to +\infty}\left(\frac{x^3+1}{x^2-2}+ax+b\right)=\lim\limits_{x\to +\infty}\frac{x^3(a+1)+bx^2-2ax+(1-2b)}{x^2-2}\)
Nếu $a\neq -1$ thì bậc của tử lớn hơn bậc của mẫu nên giới hạn tiến vô cùng chứ không phải hữu hạn $(10)$
Do đó $a=-1$
Khi đó: \(\lim\limits_{x\to +\infty}(\frac{x^3+1}{x^2-2}+ax+b)=\lim\limits_{x\to +\infty}\frac{bx^2+2x+(1-2b)}{x^2-2}=\lim\limits_{x\to +\infty}\frac{b+\frac{2}{x}+\frac{1-2b}{x^2}}{1-\frac{2}{x^2}}=b\)
Do đó $b=10$.
a: \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+x+2}}{x-1}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x}+\dfrac{2}{x^2}}}{1-\dfrac{1}{x}}=\dfrac{\sqrt{1+0+0}}{1-0}\)
\(=\dfrac{1}{1}\)
=1
b: \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{4x^2-x}+2x\right)\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{4x^2-x-4x^2}{\sqrt{4x^2-x}-2x}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x}{\sqrt{x^2\left(4-\dfrac{1}{x}\right)}-2x}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x}{-x\sqrt{4-\dfrac{1}{x}}-2x}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{1}{\sqrt{4-\dfrac{1}{x}}+2}=\dfrac{1}{\sqrt{4}+2}=\dfrac{1}{2+2}=\dfrac{1}{4}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\left(1-\dfrac{1}{x}\right)^2\left(2+\dfrac{3}{x^2}\right)}{\dfrac{4}{x^4}-1}=\dfrac{2}{-1}=-2\)
a) 
(x4 – x2 + x - 1) = x4(1 - 
) = +∞.
b) 
(-2x3 + 3x2 -5 ) = x3(-2 + 
) = +∞.
c) 
= 
= +∞.
d) \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+1}+x}{5-2x}=\lim\limits_{x\rightarrow+\infty}\dfrac{\left|x\right|\sqrt{1+\dfrac{1}{x^2}}+x}{5-2x}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{1+\dfrac{1}{x^2}}+x}{5-2x}\)\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x^2}}+1}{\dfrac{5}{x}-2}=-1\).
a: \(\lim\limits_{x\rightarrow+\infty}\left[x\left(\sqrt{x^2+2}-x\right)\right]\)
\(=\lim\limits_{x\rightarrow+\infty}\left[x\cdot\dfrac{x^2+2-x^2}{\sqrt{x^2+2}+x}\right]\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{2x}{\sqrt{x^2+2}+x}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{2}{\sqrt{1+\dfrac{2}{x^2}}+1}=\dfrac{2}{1+1}=\dfrac{2}{2}=1\)
b: \(\lim\limits_{x\rightarrow-\infty}\dfrac{3x^2-4x+6}{x-2}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2\left(3-\dfrac{4}{x}+\dfrac{6}{x^2}\right)}{x\left(1-\dfrac{2}{x}\right)}\)
\(=\lim\limits_{x\rightarrow-\infty}\left[x\cdot\dfrac{3-\dfrac{4}{x}+\dfrac{6}{x^2}}{1-\dfrac{2}{x}}\right]\)
\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow-\infty}x=-\infty\\\lim\limits_{x\rightarrow-\infty}\dfrac{3-\dfrac{4}{x}+\dfrac{6}{x^2}}{1-\dfrac{2}{x}}=\dfrac{3-0+0}{1-0}=\dfrac{3}{1}=3>0\end{matrix}\right.\)
Da nan roi mang meo lam mat het bai -.-
1/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{\dfrac{3x^3}{x^3}+\dfrac{1}{x^3}}+\sqrt{\dfrac{2x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}}{-\sqrt[4]{\dfrac{4x^4}{x^4}+\dfrac{2}{x^4}}}=\dfrac{-\sqrt[3]{3}-\sqrt{2}}{\sqrt[4]{4}}\)
2/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{8x^7}{\left(-2x^7\right)}=-\dfrac{8}{2^7}\)
3/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(4x^2-3x+4-4x^2\right)\left(\sqrt{x^2+x+1}+x\right)}{\left(x^2+x+1-x^2\right)\left(\sqrt{4x^2-3x+4}+2x\right)}=\dfrac{-3.2}{2}=-3\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{-a\sqrt{1+\dfrac{1}{x^2}}+\dfrac{2017}{x}}{1+\dfrac{2018}{x}}=-a\Rightarrow a=-\dfrac{1}{2}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{bx+1}{\sqrt{x^2+bx+1}+x}=\lim\limits_{x\rightarrow+\infty}\dfrac{b+\dfrac{1}{x}}{\sqrt{1+\dfrac{b}{x}+\dfrac{1}{x^2}}+1}=\dfrac{b}{2}=2\Rightarrow b=4\)
\(\Rightarrow P=2\)
Lời giải:
a) \(\lim\limits_{x\to -\infty}\frac{x+3}{3x-1}=\lim\limits_{x\to -\infty}\frac{1+\frac{3}{x}}{3-\frac{1}{x}}=\frac{1}{3}\)
b)
\(\lim\limits_{x\to +\infty}\frac{(\sqrt{x^2+1}+x)^n-(\sqrt{x^2+1}-x)^n}{x}=\lim\limits_{x\to +\infty} 2[(\sqrt{x^2+1}+x)^{n-1}+(\sqrt{x^2+1}+x)^{n-1}(\sqrt{x^2+1}-x)+....+(\sqrt{x^2+1}-x)^{n-1}]\)
\(=+\infty\)
\(\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\left(a+1\right)x^2-\left(2a+b\right)x+2b+1}{x-2}\right)\)
Giới hạn hữu hạn khi \(a+1=0\Rightarrow a=-1\)
Khi đó: \(\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\left(2-b\right)x+2b+1}{x-2}\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{2-b+\dfrac{2b+1}{x}}{1-\dfrac{2}{x}}=2-b=-5\)
\(\Rightarrow b=7\)