lim x->0\(\frac{\sqrt{1-2x}-1}{\sqrt{5x+4}-2}\) ( giúp em với ạ )
CN
Những câu hỏi liên quan
1) lim \(\frac{-x^2+3x}{x^3-2x^2+x}\) (x->1)
2) lim \(\frac{\sqrt{1+2x}-\sqrt[3]{1+3x}}{x^2}\) (x->0)
3) lim \(\frac{x\sqrt[3]{x^3+1
}}{2-x\sqrt{1+4x^2}}\) (x-> âm vô cùng )
4) lim \(\frac{\cos^9x-1}{x}\) (x->0)
giúp mình với ạ
tính \(\lim\limits_{x\rightarrow1}\dfrac{2-\sqrt{2x-1}.\sqrt[3]{5x+3}}{x-1}\) mn giúp mk với ạ
\(\lim\limits_{x\rightarrow1}\dfrac{2-\sqrt[]{2x-1}\sqrt[3]{5x+3}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{2-2\sqrt[]{2x-1}+2\sqrt[]{2x-1}-\sqrt[]{2x-1}.\sqrt[3]{5x+3}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{2\left(1-\sqrt[]{2x-1}\right)+\sqrt[]{2x-1}\left(2-\sqrt[3]{5x+3}\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{-\dfrac{4\left(x-1\right)}{1+\sqrt[]{2x-1}}-\dfrac{5\sqrt[]{2x-1}\left(x-1\right)}{4+2\sqrt[3]{5x+3}+\sqrt[3]{\left(5x+3\right)^2}}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\left(-\dfrac{4}{1+\sqrt[]{2x-1}}-\dfrac{5\sqrt[]{2x-1}}{4+2\sqrt[3]{5x+3}+\sqrt[3]{\left(5x+3\right)^2}}\right)\)
\(=-\dfrac{4}{1+1}-\dfrac{5\sqrt[]{1}}{4+4+4}=-\dfrac{29}{12}\)
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Tìm các giới hạn sau:
a) limlimits_{xrightarrow-1}frac{sqrt[3]{x}+1}{2x^2+5x+3}
b) limlimits_{xrightarrow1}frac{sqrt[3]{x^2}-2sqrt[3]{x}+1}{left(x-1right)^2}
c)limlimits_{xrightarrow1}frac{sqrt[4]{x}-1}{x^3+x^2-2}
d) limlimits_{xrightarrow-2}frac{sqrt[3]{2x+12}+x}{x^2+2x}
mọi người ơi giúp mình với, mình cảm ơn nhiều ạ :((((
Đọc tiếp
Tìm các giới hạn sau:
a) \(\lim\limits_{x\rightarrow-1}\frac{\sqrt[3]{x}+1}{2x^2+5x+3}\)
b) \(\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{\left(x-1\right)^2}\)
c)\(\lim\limits_{x\rightarrow1}\frac{\sqrt[4]{x}-1}{x^3+x^2-2}\)
d) \(\lim\limits_{x\rightarrow-2}\frac{\sqrt[3]{2x+12}+x}{x^2+2x}\)
mọi người ơi giúp mình với, mình cảm ơn nhiều ạ :((((
Lời giải:
a)
\(\lim\limits_{x\to-1}\frac{\sqrt[3]{x}+1}{2x^2+5x+3}=\lim\limits_{x\to-1}\frac{x+1}{\left(\sqrt[3]{x^2}-\sqrt[3]{x}+1\right)\left(x+1\right)\left(2x+3\right)}\)
\(\lim\limits_{x\to-1}\frac{1}{\left(\sqrt[3]{x^2}-\sqrt[3]{x}+1\right)\left(2x+3\right)}=\frac{1}{\left(\sqrt[3]{\left(-1\right)^2}-\sqrt[3]{-1}+1\right)\left(2.-1+3\right)}=\frac{1}{3}\)
b)
\(\lim\limits_{x\to1}\frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(\sqrt[3]{x}-1\right)^2}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(x-1\right)^2}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2\left(x-1\right)^2}\)
\(=\lim\limits_{x\to1}\frac{1}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2}=\frac{1}{\left(1+1+1\right)^2}=\frac{1}{9}\)
c)
\(\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{x^3+x^2-2}=\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{(x-1)(x^2+2x+2)}=\lim_{x\to 1}\frac{x-1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x-1)(x^2+2x+2)}\)
\(=\lim_{x\to 1}\frac{1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x^2+2x+2)}=\frac{1}{(1+1)(1+1)(1+2.1+2)}=\frac{1}{20}\)
d)
\(\lim_{x\to -2}\frac{\sqrt[3]{2x+12}+x}{x^2+2x}=\lim_{x\to -2}\frac{2x+12+x^3}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}\)
\(=\lim_{x\to -2}\frac{(x+2)(x^2-2x+6)}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}=\lim_{x\to -2}\frac{x^2-2x+6}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x}\)
\(=\frac{-7}{12}\)
Lời giải:
a)
\(\lim\limits_{x\to-1}\frac{\sqrt[3]{x}+1}{2x^2+5x+3}=\lim\limits_{x\to-1}\frac{x+1}{\left(\sqrt[3]{x^2}-\sqrt[3]{x}+1\right)\left(x+1\right)\left(2x+3\right)}\)
\(\lim\limits_{x\to-1}\frac{1}{\left(\sqrt[3]{x^2}-\sqrt[3]{x}+1\right)\left(2x+3\right)}=\frac{1}{\left(\sqrt[3]{\left(-1\right)^2}-\sqrt[3]{-1}+1\right)\left(2.-1+3\right)}=\frac{1}{3}\)
b)
\(\lim\limits_{x\to1}\frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(\sqrt[3]{x}-1\right)^2}{\left(x-1\right)^2}=\lim\limits_{x\to1}\frac{\left(x-1\right)^2}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2\left(x-1\right)^2}\)
\(=\lim\limits_{x\to1}\frac{1}{\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)^2}=\frac{1}{\left(1+1+1\right)^2}=\frac{1}{9}\)
c)
\(\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{x^3+x^2-2}=\lim_{x\to 1}\frac{\sqrt[4]{x}-1}{(x-1)(x^2+2x+2)}=\lim_{x\to 1}\frac{x-1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x-1)(x^2+2x+2)}\)
\(=\lim_{x\to 1}\frac{1}{(\sqrt{x}+1)(\sqrt[4]{x}+1)(x^2+2x+2)}=\frac{1}{(1+1)(1+1)(1+2.1+2)}=\frac{1}{20}\)
d)
\(\lim_{x\to -2}\frac{\sqrt[3]{2x+12}+x}{x^2+2x}=\lim_{x\to -2}\frac{2x+12+x^3}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}\)
\(=\lim_{x\to -2}\frac{(x+2)(x^2-2x+6)}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x(x+2)}=\lim_{x\to -2}\frac{x^2-2x+6}{(\sqrt[3]{(2x+12)^2}-x\sqrt[3]{2x+12}+x^2).x}\)
\(=\frac{-7}{12}\)
1/ lim x-> +∞ \(\dfrac{1}{\sqrt{x^2+x+1}-x}\)
2/ lim x->+∞ \(\dfrac{3x-2\sqrt{x}+\sqrt{x^4-5x}}{2x^2+4x-5}\)
help mình vs ạ. cảm ơn ~~~~
a/ ^{lim}_{x-0}frac{sqrt{1+x}-sqrt[3]{1+x}}{x}
b/^{lim}_{x-1}left(frac{1}{1-x}-frac{1}{1-x^3}right)
c/ ^{lim}_{x-+infty}left(sqrt[3]{2x-1}-sqrt[3]{2x+1}right)
d/ ^{lim}_{x--infty}left(sqrt[3]{3x^3-1}+sqrt{x^2+2}right)
e/^{lim}_{x-2}left(frac{1}{x^2-3x+2}+frac{1}{x^2-5x+6}right)
f/ ^{lim}_{x-0^{+-}}left(frac{2x}{sqrt{4x^2+x^3}}right)
Đọc tiếp
a/ \(^{lim}_{x->0}\frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}\)
b/\(^{lim}_{x->1}\left(\frac{1}{1-x}-\frac{1}{1-x^3}\right)\)
c/ \(^{lim}_{x->+\infty}\left(\sqrt[3]{2x-1}-\sqrt[3]{2x+1}\right)\)
d/ \(^{lim}_{x->-\infty}\left(\sqrt[3]{3x^3-1}+\sqrt{x^2+2}\right)\)
e/\(^{lim}_{x->2}\left(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}\right)\)
f/ \(^{lim}_{x->0^{+-}}\left(\frac{2x}{\sqrt{4x^2+x^3}}\right)\)
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.
\(=\frac{2x}{x\sqrt{4+x}}=\frac{2}{\sqrt{4+x}}=1\)
Bài 1
a. limlimits_{xrightarrow-infty}frac{sqrt{4x^2}+1}{3x-1}
b. limlimits_{xrightarrow+infty}frac{sqrt{9x^2+x+1}-sqrt{4x^2+2x+1}}{x+1}
c. limlimits_{xrightarrow+infty}frac{sqrt{x+2x+3}+4x+1}{sqrt{4x^2+1}+2-x}
d. limlimits_{xrightarrow+infty}frac{3x-2sqrt{x}+sqrt{x^4-5x}}{2x^2+4x-5}
Bài 2
a. limlimits_{xrightarrow-infty}frac{2x+1}{x-1}
b. limlimits_{xrightarrow-infty}frac{2x^3+3}{x^3-2x^2+1}
c. limlimits_{xrightarrow+infty}frac{left(3x^2+1right)left(5x+3right)}{left(2x^3-1r...
Đọc tiếp
Bài 1
a. \(\lim\limits_{x\rightarrow-\infty}\frac{\sqrt{4x^2}+1}{3x-1}\)
b. \(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9x^2+x+1}-\sqrt{4x^2+2x+1}}{x+1}\)
c. \(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+2x+3}+4x+1}{\sqrt{4x^2+1}+2-x}\)
d. \(\lim\limits_{x\rightarrow+\infty}\frac{3x-2\sqrt{x}+\sqrt{x^4-5x}}{2x^2+4x-5}\)
Bài 2
a. \(\lim\limits_{x\rightarrow-\infty}\frac{2x+1}{x-1}\)
b. \(\lim\limits_{x\rightarrow-\infty}\frac{2x^3+3}{x^3-2x^2+1}\)
c. \(\lim\limits_{x\rightarrow+\infty}\frac{\left(3x^2+1\right)\left(5x+3\right)}{\left(2x^3-1\right)\left(x+4\right)}\)
Bài 1:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)
Bài 2:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)
\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)
Tính giới hạn
a) \(\lim\limits_{x->0}\dfrac{\sqrt[m]{2x+1}-1}{\sqrt[n]{x+1}-1}\)
b) \(\lim\limits_{x->3}\dfrac{\sqrt[4]{5x+1}-2}{x-3}\)
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.
a/ \(\lim\limits_{x\rightarrow-1}\dfrac{2x^3-5x-4}{\left(x+1\right)^2}=\dfrac{2.\left(-1\right)^3-5\left(-1\right)-4}{\left(-1+1\right)^2}=-\dfrac{1}{0}=-\infty\)
b/ \(\lim\limits\left(x^3+2\sqrt{x^5}-1\right)=\lim\limits x^3\left(1+0-0\right)=+\infty\)
Đúng 2
Bình luận (2)
a,^{lim}_{x-2}frac{sqrt[3]{8x+11}-sqrt{x+7}}{x^2-3x+2}
b, ^{lim}_{x-0}frac{2sqrt{1+x}-sqrt[3]{8-x}}{x}
c, ^{lim}_{x-1}frac{sqrt{5-x^3}-sqrt[3]{x^2+7}}{x^2-1}
d,^{lim}_{x-0}frac{sqrt{1+2x}.sqrt[3]{1+4x}-1}{x}
e,^{lim}_{x-1}frac{x^4-1}{x^3-2x^2+x}
f,^{lim}_{x-1}left(frac{1}{1-x}-frac{3}{1-x^3}right)
Đọc tiếp
a,\(^{lim}_{x->2}\frac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^2-3x+2}\)
b, \(^{lim}_{x->0}\frac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)
c, \(^{lim}_{x->1}\frac{\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}\)
d,\(^{lim}_{x->0}\frac{\sqrt{1+2x}.\sqrt[3]{1+4x}-1}{x}\)
e,\(^{lim}_{x->1}\frac{x^4-1}{x^3-2x^2+x}\)
f,\(^{lim}_{x->1}\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right)\)
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