\(\lim\limits_{x\rightarrow-\infty}\left(4x^5-3x^2+1\right)=\lim\limits_{x\rightarrow-\infty}x^5\left(4-\frac{3}{x^3}+\frac{1}{x^5}\right)=-\infty.4=-\infty\)
\(\lim\limits_{x\rightarrow4}\frac{1-x}{\left(x-4\right)^2}=\frac{-3}{0}=-\infty\)
Câu tiếp theo đề thiếu, ko thấy yêu cầu gì hết
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/ \(^{lim}_{x->0^{+-}}\left(\frac{2x}{\sqrt{4x^2+x^3}}\right)\)
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,\(^{lim}_{x->1}\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right)\)
Cho f(x) thỏa mãn : \(_{\lim\limits_{x\rightarrow-1}\dfrac{2f\left(x\right)+1}{x+1}=5}\)
Tính I= \(\lim\limits_{x\rightarrow-1}\dfrac{\left(4f\left(x\right)+3\right)\left(\sqrt{4f\left(x\right)^2+2f\left(x\right)+4}\right)-2}{x^2-1}\)
1, \(\lim\limits_{x\rightarrow1}\frac{2x^2-3x+1}{x^3-x^2-x+1}\)
2, \(\lim\limits_{x\rightarrow2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}\)
3, \(\lim\limits_{x\rightarrow0}\frac{1-\sqrt[3]{x-1}}{x}\)
4, \(\lim\limits_{x\rightarrow-\infty}\frac{x^2-5x+1}{x^2-2}\)
5, \(\lim\limits_{x\rightarrow+\infty}\frac{2x^2-4}{x^3+3x^2-9}\)
6, \(\lim\limits_{x\rightarrow2^-}\frac{2x-1}{x-2}\)
7, \(\lim\limits_{x\rightarrow3^+}\frac{8+x-x^2}{x-3}\)
8, \(\lim\limits_{x\rightarrow-\infty}\left(8+4x-x^3\right)\)
9, \(\lim\limits_{x\rightarrow-1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}\)
10, \(\lim\limits_{x\rightarrow-\infty}\frac{\left(2x^2+1\right)^2\left(5x+3\right)}{\left(2x^3-1\right)\left(x+1\right)^2}\)
11, \(\lim\limits_{x\rightarrow-\infty}\frac{\sqrt{x^2+2x}}{x+3}\)
12, \(\lim\limits_{x\rightarrow1}\frac{\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}\)
13, \(\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{x+1}+\sqrt{x+4}-3}{x}\)
14, \(\lim\limits_{x\rightarrow0}\frac{\left(x^2+2020\right)\sqrt{1+3x}-2020}{x}\)
15, \(\lim\limits_{x\rightarrow+\infty}\left(2x-\sqrt{4x^2-3}\right)\)
16, \(\lim\limits_{x\rightarrow a}\frac{x^2-\left(a+1\right)x+a}{x^3-a^3}\)
17, \(\lim\limits_{x\rightarrow1}\frac{x^n-nx+n-1}{\left(x-1\right)^2}\)
18, \(f\left(x\right)=\left\{{}\begin{matrix}\frac{x^2-2x}{8-x^3}\\\frac{x^4-16}{x-2}\end{matrix}\right.\) khi x>2,khi x<2 tại x=2
Cho \(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-2x+1}{x-1}=3\)
Tính \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{3f\left(x\right)+1}-x-1}{\sqrt{4x+5}-3x-2}\)
Bài 1 tìm các giới hạn sau :
a, lim 2x²-3x-2/x-2
(limx->2)
b, lim x³-3x²+5x-3/x²-1
(limx->1)
c, limx²+2x/x²+4x+4
(limx->2)
d, limx³-x²-x+1/x²-3x+2
(lim x->1)
e, limx³-5x²+3x+9/x4-8x²-9
(lim x->1)
f, lim x4-1/x³-2x²+3
(limx->-1)
g, limx²+2x-3/2x²-x-1
(limx->1)
h,lim x³-3x+2/4-x²
(lim x->-2)
i, lim4x6-5x5+1/x²-1
(lim x->1)
k, lim x mũ m -1/ x mũ n -1
(lim x->1)m, n thuộc N
tìm các giới hạn sau:
a, \(\lim\limits_{x\rightarrow1}\frac{x^4-1}{x^3-2x^2+1}\) ( câu a,b chỉ cần thay số vào thôi đúng k ạ nếu là thay số thì k cần trình bày nữa đâu )
b, \(\lim\limits_{x\rightarrow-1}\frac{x^5+1}{x^3+1}\)
c, \(\lim\limits_{x\rightarrow3}\frac{x^3-5x^2+3x+9}{x^4-8x^2-9}\)
d, \(\lim\limits_{x\rightarrow1}\frac{x-5x^5+4x^6}{\left(1-x\right)^2}\)
e, \(\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}\)
f, \(\lim\limits_{x\rightarrow-2}\frac{x^4-16}{x^3+2x^2}\)
Tính các giới hạn sau đây :
\(L_1=lim\frac{x^3+3x^2-2x}{x^5+4x}\left(x\rightarrow0\right)\)
\(L_2=lim\frac{x^3-3x+2}{\left(4-2x\right)^3}\left(x\rightarrow+\infty\right)\)
\(L_3=lim\frac{2x^2+3x+1}{x^2+x}\left(x\rightarrow-1\right)\)
\(L_4=lim\frac{x^2-4x+1}{4-x^2}\left(x\rightarrow2\right)\)
\(L_5=lim\frac{\sqrt{x+1}-2}{x-2}\left(x\rightarrow3\right)\)
\(L_6=lim\frac{\sqrt{x+3}-x-1}{x^2-1}\left(x\rightarrow1\right)\)
\(L_7=lim\left(\sqrt{x^2+x+1}-x+1\right)\left(x\rightarrow+\infty\right)\)
\(L_8=lim\left(\sqrt{x^2+x+1}-3x+2\right)\left(x\rightarrow-\infty\right)\)
Tính các giới hạn sau:
Câu 1:
a, limx→\(\pm\)∞ \(\dfrac{\left(2x-3\right)^2\left(4x+7\right)^3}{\left(3x-4\right)^2\left(5x^2-1\right)}\)
b, limx→\(\pm\)∞ \(\dfrac{\sqrt[3]{x^3+2x^2+x}}{2x-2}\)
c, limx→\(\pm\)∞ \(\dfrac{\sqrt[3]{\left(x^3+2x^2\right)^2}+x^3\sqrt{x^3+2x^2}+x^2}{3x^2-2x}\)
d, limx→+∞ \(\dfrac{\left(2-3x\right)^3\left(x+1\right)^2}{1-4x^5}\)
e, limx→+∞ \(\dfrac{\left(2x-3\right)^{20}\left(3x+2\right)^{20}}{\left(2x+1\right)^{50}}\)
g, limx→+∞ \(\dfrac{\left(2x-3\right)^3\left(4x^5+7\right)^9}{11x^{47}-8}\)
