a) lim \(\dfrac{2x^3-5x-4}{\left(x+1\right)^2}\) khi x tiến đến -1.
b) lim (x3 + 2x2\(\sqrt{x}\) - 1) khi x tiến đến dương vô cùng.
Giúp mình với ạ.
tìm lim (2x^2 -xsinx+1)/(x^2 -xcosx +2) (x tiến đến dương vô cùng)
lim (cos10x+xsin10x)/(x căn x +2) (x tiến đến âm vô cùng)
a) lim \(\dfrac{x\sqrt{x^2+1}-2x+1}{^3\sqrt{2x^3-2}+1}\)
x-> -∞
b) lim \(\dfrac{\left(2x+1\right)^3\left(x+2\right)^4}{\left(3-2x\right)^7}\)
x-> -∞
c) lim \(\dfrac{\sqrt{4x^2+x}+^3\sqrt{8x^3+x-1}}{^4\sqrt{x^4+3}}\)
x-> +∞
a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x\sqrt{x^2+1}}{x}-\dfrac{2x}{x}+\dfrac{1}{x}}{\sqrt[3]{\dfrac{2x^3}{x^3}-\dfrac{2x}{x^3}}+\dfrac{1}{x}}=0\)
b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{8x^7}{x^7}}{\dfrac{\left(-2x^7\right)}{x^7}}=-\dfrac{8}{2^7}\)
c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)
Tìm giới hạn
1) \(\xrightarrow[x->3]{lim}\dfrac{x^2-5x+6}{\sqrt{2x+3}-\sqrt{4x-3}}\)
2) \(\xrightarrow[x->1]{lim}\dfrac{\sqrt{x^2+2}-\sqrt{4x-1}}{x-1}\)
3) \(\xrightarrow[x->-1]{lim}\dfrac{x-2}{x\left|x+1\right|}\)
4) \(\xrightarrow[x->a]{lim}\dfrac{x^n-a^n}{x-a}\)
5) \(\xrightarrow[x->1]{lim}(\dfrac{n}{1-x^n}-\dfrac{1}{1-x})\)
6) \(\xrightarrow[x->1]{lim}\dfrac{x^n-nx+n-1}{\left(x-1\right)^2}\)
Giá trị của các giới hạn :
a, lim\(\left(\sqrt[3]{3x^3-1}+\sqrt{x^2+1}\right)\) khi x→\(-\infty\)
b, lim\(\left(\sqrt{x^2+x}-\sqrt[3]{x^3-x^2}\right)\) khi x→\(+\infty\)
c, lim\(\left(\sqrt[3]{2x-1}-\sqrt[3]{2x+1}\right)\) khi x→\(+\infty\)
a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}-x}+\lim\limits_{x\rightarrow-\infty}\dfrac{3x^3-1-x^3}{\sqrt[3]{\left(3x^3-1\right)^2}+x\sqrt[3]{3x^3-1}+x^2}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}}{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}-\dfrac{x}{x}}+\lim\limits_{x\rightarrow-\infty}\dfrac{-\dfrac{1}{x^2}}{\dfrac{\sqrt[3]{\left(3x^3-1\right)^2}}{x^2}+\dfrac{x\sqrt[3]{3x^3-1}}{x^2}+\dfrac{x^2}{x^2}}=0\)
b/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+x-x^2}{\sqrt{x^2+x}+x}+\lim\limits_{x\rightarrow+\infty}\dfrac{x^3-x^3+x^2}{x^2+x\sqrt[3]{x^3-x^2}+\sqrt[3]{\left(x^3-x^2\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}}+\dfrac{x}{x}}+\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x^2}{x^2}}{\dfrac{x^2}{x^2}+\dfrac{x\sqrt[3]{x^3-x^2}}{x^2}+\dfrac{\sqrt[3]{\left(x^3-x^2\right)^2}}{x^2}}\)
\(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{2x-1-2x-1}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{4x^2-1}+\sqrt[3]{\left(2x+1\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{2}{x^{\dfrac{2}{3}}}}{\dfrac{\sqrt[3]{\left(2x-1\right)^2}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{4x^2-1}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{\left(2x+1\right)^2}}{x^{\dfrac{2}{3}}}}=0\)
Check lai ho minh nhe :v
1/ \(\lim\limits_{x\to 1}\) \(\dfrac{\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1}\)
2/ \(\lim\limits_{x \to \ +\infty} \)\(x\left[\sqrt{4x^2+5}-\sqrt[3]{8x^3-1}\right]\)
3/ \(\lim\limits_{x\to 1}\)\(\dfrac{x^3-2x-1}{x^5-2x-1}\)
Giải giúp mình với ạ
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt[3]{7+x^3}-2\right)-\left(\sqrt{3+x^2}-2\right)}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^3-1}{\left(\sqrt[3]{7+x^3}\right)^2+2\sqrt[3]{7+x^3}+4}-\dfrac{x^2-1}{\sqrt{3+x^2}+2}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^2+x+1}{\left(\sqrt[3]{7+x^3}\right)^2+2\sqrt[3]{7+x^3}+4}-\dfrac{x+1}{\sqrt{3+x^2}+2}}{1}=\dfrac{3}{12}-\dfrac{2}{4}=\dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{4}\).
Tìm các giới hạn sau :
a, lim\(\dfrac{2x^2+x-6}{x^3+8}\) khi x→-2
b, lim\(\dfrac{x^4-x^2-72}{x^2-2x-3}\) khi x→3
c, lim\(\dfrac{x^5+1}{x^3+1}\) khi x→-1
d, lim \(\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)\) khi x→1
a) \(\lim\limits_{x\rightarrow-2}\dfrac{2x^2+x-6}{x^3+8}=\lim\limits_{x\rightarrow-2}\dfrac{\left(2x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\\ =\lim\limits_{x\rightarrow-2}\dfrac{2x-3}{x^2-2x+4}=-\dfrac{7}{12}\).
b) \(\lim\limits_{x\rightarrow3}\dfrac{x^4-x^2-72}{x^2-2x-3}=\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\\ =\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)}{x+1}=\dfrac{51}{2}\).
c) \(\lim\limits_{x\rightarrow-1}\dfrac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\lim\limits_{x\rightarrow-1}\dfrac{x^4-x^3+x^2-x+1}{x^2-x+1}=\dfrac{5}{3}\).
d) \(\lim\limits_{x\rightarrow1}\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\\ =\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{x+1}=-\dfrac{1}{2}\).
Giúp mình với ạ
1) lim\(\dfrac{x-5x^2+1}{x^2-1}\)(x-->-∞)
2) lim\(\dfrac{5x^3\left(2-x^2\right)^3\left(4x^2+1\right)^2}{4x^{13}+x^2-6}\)(x-->+∞)
3) lim\(\dfrac{4x-\sqrt{9x^2+x}}{3-x}\)(x-->+∞)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{x-5x^2+1}{x^2-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}-5+\dfrac{1}{x^2}}{1-\dfrac{1}{x^2}}=\dfrac{-5}{1}=-5\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{5x^3\left(2-x^2\right)^3\left(4x^2+1\right)^2}{4x^{13}+x^2-6}=\lim\limits_{x\rightarrow+\infty}\dfrac{5\left(\dfrac{2}{x^2}-1\right)^3\left(4+\dfrac{1}{x^2}\right)^2}{4+\dfrac{1}{x^{11}}-\dfrac{6}{x^{13}}}=\dfrac{5.\left(-1\right)^3.4^2}{4}=-20\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{4x-\sqrt{9x^2+x}}{3-x}=\lim\limits_{x\rightarrow+\infty}\dfrac{4-\sqrt{9+\dfrac{1}{x}}}{\dfrac{3}{x}-1}=\dfrac{4-3}{-1}=-1\)
Giúp mình với ạ
1) lim\(\dfrac{3x^2+5}{x^3-x+2}\)(x-->+∞)
2) lim\(\dfrac{2x^2\left(3x^2-5\right)^3\left(1-x\right)^5}{3x^{14}+x^2-1}\)(x-->-∞)
3) lim\(\dfrac{3x-\sqrt{2x^2+5}}{x^2-4}\)(x-->+∞)
1 ) \(lim_{x\rightarrow+\infty}\dfrac{3x^2+5}{x^3-x+2}=lim_{x\rightarrow+\infty}\dfrac{\dfrac{3}{x}+\dfrac{5}{x^3}}{1-\dfrac{1}{x^2}+\dfrac{2}{x^3}}=0\)
2 ) \(lim_{x\rightarrow-\infty}\dfrac{2x^2\left(3x^2-5\right)^3\left(1-x\right)^5}{3x^{14}+x^2-1}\) \(=lim_{x\rightarrow-\infty}\dfrac{\dfrac{2}{x}\left(3-\dfrac{5}{x^2}\right)^3\left(\dfrac{1}{x}-1\right)^5}{3+\dfrac{1}{x^{12}}-\dfrac{1}{x^{14}}}=0\)
3 ) \(lim_{x\rightarrow+\infty}\dfrac{3x-\sqrt{2x^2+5}}{x^2-4}=lim_{x\rightarrow+\infty}\dfrac{\left(7x^2-5\right)}{\left(3x+\sqrt{2x^2+5}\right)\left(x^2-4\right)}\)
\(=lim_{x\rightarrow+\infty}\dfrac{\dfrac{7}{x}-\dfrac{5}{x^3}}{\left(3+\sqrt{2+\dfrac{5}{x^2}}\right)\left(1-\dfrac{4}{x^2}\right)}=0\)
tìm giới hạn
a.\(\lim\limits_{x\rightarrow-1}\left(\dfrac{2x^2-x-3}{x^3+1}\right)\)
b.\(\lim\limits_{x\rightarrow3}\left(\dfrac{2x^3-5x^2-2x-3}{\sqrt[3]{x+5}-2}\right)\)
Dạng 0/0 một là phân tích đa thức thành nhân tử để rút gọn mẫu khỏi dạng 0/0. Hoặc là nhân liên hợp
a/ \(=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x-\dfrac{3}{2}\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{x}{x^2}-\dfrac{3}{2x^2}}{\dfrac{x^2}{x^2}-\dfrac{x}{x^2}+\dfrac{1}{x^2}}=0\)
b/ \(=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(2x^2+x+1\right)\left[\left(\sqrt[3]{x+5}\right)^2+2\sqrt[3]{x+5}+4\right]}{x-3}\)
\(=\left(2.3^2+3+1\right)\left[\left(\sqrt[3]{3+5}\right)^2+2\sqrt[3]{3+5}+4\right]=...\)
bn nên đăng ở môn cần nha!