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\(\lim\limits_{x\rightarrow3}f\left(x\right)=\lim\limits_{x\rightarrow3}\frac{8x^{2016}-24x^{2015}}{x^{2017}+2x^{2016}-15x^{2015}}=\lim\limits_{x\rightarrow3}\frac{8\left(x-3\right)}{x^2+2x-15}=\lim\limits_{x\rightarrow3}\frac{8\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}=\lim\limits_{x\rightarrow3}\frac{8}{x+5}=1\)
\(\lim\limits_{x\rightarrow1}g\left(x\right)=\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+2}-2+2-\sqrt{3x+1}}{m\left(x-1\right)\left(x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{\frac{2\left(x-1\right)}{\sqrt{2x+2}+2}-\frac{3\left(x-1\right)}{2+\sqrt{3x+1}}}{m\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{2}{\sqrt{2x+2}+2}-\frac{3}{2+\sqrt{3x+1}}}{m\left(x+1\right)}=\frac{\frac{2}{4}-\frac{3}{4}}{2m}=-\frac{1}{8m}\)
\(\Rightarrow-\frac{1}{8m}=1\Rightarrow m=-\frac{1}{8}\)
\(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x^2+1}+x}{3x+5}=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{1}{x^2}}+1}{3+\frac{5}{x}}=\frac{2}{3}\)
Lời giải:
\(\lim\limits _{x\to +\infty}\sqrt{\frac{3x^4+4x^5+2}{9x^5+5x^4+4}}=\lim\limits _{x\to +\infty}\sqrt{\frac{\frac{3}{x}+4+\frac{2}{x^5}}{9+\frac{5}{x}+\frac{4}{x^5}}}=\sqrt{\frac{4}{9}}=\frac{2}{3}\)
Đáp án B.
\(a=\lim\limits_{x\rightarrow0}\frac{x^2}{x\left(\sqrt{1+x^2}+1\right)}=\lim\limits_{x\rightarrow0}\frac{x}{\sqrt{1+x^2}+1}=\frac{0}{2}=0\)
\(b=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+7}-2+2-\sqrt{5-x^2}}{x-1}=\lim\limits_{x\rightarrow1}\frac{\frac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{\left(x-1\right)\left(x+1\right)}{2+\sqrt{5-x^2}}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\left(\frac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{x+1}{2+\sqrt{5-x^2}}\right)=\frac{1}{12}+\frac{1}{2}=\frac{7}{12}\)
\(c=\lim\limits_{x\rightarrow0}\frac{2x}{x\left(\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{\left(1+x\right)\left(1-x\right)}+\sqrt[3]{\left(1-x\right)^2}\right)}=\lim\limits_{x\rightarrow0}\frac{2}{\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{\left(1+x\right)\left(1-x\right)}+\sqrt[3]{\left(1-x\right)^2}}=\frac{2}{3}\)
\(d=\frac{\sqrt[3]{6}}{0}=+\infty\)
a) 
(x4 – x2 + x - 1) = x4(1 - 
) = +∞.
b) 
(-2x3 + 3x2 -5 ) = x3(-2 + 
) = +∞.
c) 
= 
= +∞.
d) \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+1}+x}{5-2x}=\lim\limits_{x\rightarrow+\infty}\dfrac{\left|x\right|\sqrt{1+\dfrac{1}{x^2}}+x}{5-2x}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{1+\dfrac{1}{x^2}}+x}{5-2x}\)\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x^2}}+1}{\dfrac{5}{x}-2}=-1\).
Bạn tự hiểu là giới hạn tiến đến đâu nhé, làm biếng gõ đủ công thức
a. \(\frac{\sqrt{1+x}-1+1-\sqrt[3]{1+x}}{x}=\frac{\frac{x}{\sqrt{1+x}+1}-\frac{x}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}}{x}=\frac{1}{\sqrt{1+x}+1}-\frac{1}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)
b.
\(\frac{1-x^3-1+x}{\left(1-x\right)^2\left(1+x+x^2\right)}=\frac{x\left(1-x\right)\left(1+x\right)}{\left(1-x\right)^2\left(1+x+x^2\right)}=\frac{x\left(1+x\right)}{\left(1-x\right)\left(1+x+x^2\right)}=\frac{2}{0}=\infty\)
c.
\(=\frac{-2}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{\left(2x+1\right)^2}+\sqrt[3]{\left(2x-1\right)\left(2x+1\right)}}=\frac{-2}{\infty}=0\)
d.
\(=x\sqrt[3]{3-\frac{1}{x^3}}-x\sqrt{1+\frac{2}{x^2}}=x\left(\sqrt[3]{3-\frac{1}{x^3}}-\sqrt{1+\frac{2}{x^2}}\right)=-\infty\)
e.
\(=\frac{2x^2-8x+8}{\left(x-1\right)\left(x-2\right)\left(x-2\right)\left(x-3\right)}=\frac{2\left(x-2\right)^2}{\left(x-1\right)\left(x-3\right)\left(x-2\right)^2}=\frac{2}{\left(x-1\right)\left(x-3\right)}=\frac{2}{-1}=-2\)
f.
\(=\frac{2x}{x\sqrt{4+x}}=\frac{2}{\sqrt{4+x}}=1\)
Ta có:
Chọn D