tìm giới hạn : \(\lim\limits_{x\rightarrow1}\frac{x^4+x^3-2}{x^5-x^2}\)
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Những câu hỏi liên quan
AH
Akai Haruma
Giáo viên
12 tháng 3 2020
a.
\(\lim\limits_{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}=\lim_{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{(x-1)^3(3x+1)}=\lim\limits _{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{3x+1}.\lim\limits_{x\to 1+}\frac{1}{(x-1)^3}\)
\(=\frac{1}{4}.(+\infty)=+\infty \)
Hoàn toàn tương tự:
\(\lim\limits_{x\to 1-}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}=-\infty \)
Do đó: \(\lim\limits_{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\neq \lim\limits_{x\to 1-}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\) nên không tồn tại \(\lim\limits_{x\to 1}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\)
AH
Akai Haruma
Giáo viên
12 tháng 3 2020
b.
\(\lim\limits_{x\to 1+}\frac{x^3-3x^2+2}{x^4-4x+3}=\lim\limits_{x\to 1+}\frac{(x-1)(x^2-2x-2)}{(x-1)^2(x^2+2x+3)}=\lim\limits_{x\to 1+}\frac{x^2-2x-2}{(x-1)(x^2+2x+3)}\)
\(=\lim\limits_{x\to 1+}\frac{x^2-2x-2}{x^2+2x+3}.\lim\limits_{x\to 1+}\frac{1}{x-1}=\frac{-1}{2}.(+\infty)=-\infty \)
Tương tự \(\lim\limits_{x\to 1-}\frac{x^3-3x^2+2}{x^4-4x+3}=+\infty \)
Do đó không tồn tại \(\lim\limits_{x\to 1}\frac{x^3-3x^2+2}{x^4-4x+3}\)
c.
\(\lim\limits_{x\to 1}\frac{x^3-2x-1}{x^5-2x-1}=\frac{1^3-2.1-1}{1^5-2.1-1}=1\)
d.
\(\lim\limits_{x\to -1}\frac{(x+2)^2-1}{x^2-1}=\lim\limits_{x\to -1}\frac{(x+2-1)(x+2+1)}{(x-1)(x+1)}=\lim\limits_{x\to -1}\frac{x+3}{x-1}=-1\)
HA
1
HT
24 tháng 1 2021
a/ L'Hospital:
\(=\lim\limits_{x\rightarrow2}\dfrac{x-\left(x+2\right)^{\dfrac{1}{2}}}{\left(4x+1\right)^{\dfrac{1}{2}}-3}=\lim\limits_{x\rightarrow2}\dfrac{1-\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}{\dfrac{1}{2}\left(4x+1\right)^{-\dfrac{1}{2}}.4}=\dfrac{1-\dfrac{1}{2}.4^{-\dfrac{1}{2}}}{2.9^{-\dfrac{1}{2}}}=\dfrac{9}{8}\)
b/ L'Hospital:\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7\right)^{\dfrac{1}{2}}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{2}\left(2x+7\right)^{-\dfrac{1}{2}}.2+1}{3x^2-8x}=\dfrac{9^{-\dfrac{1}{2}}+1}{3-8}=-\dfrac{4}{15}\)
VT
1
4 tháng 5 2016
\(L=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)+\left(x^2-1\right)+\left(x^3-1\right)+\left(x^4-1\right)+\left(x^5-1\right)+\left(x^6-1\right)}{\left(x-1\right)+\left(x^2-1\right)+\left(x^3-1\right)+\left(x^4-1\right)+\left(x^5-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left[1+\left(x+1\right)+\left(x^2+x+1\right)+...+\left(x^5+x^4+x^3+x^2+x+1\right)\right]}{\left(x-1\right)\left[1+\left(x+1\right)+\left(x^2+x+1\right)+...+\left(x^4+x^3+x^2+x+1\right)\right]}\)
\(=\lim\limits_{1\rightarrow x}\frac{1+\left(x+1\right)+\left(x^2+x+1\right)+.....+\left(x^5+x^4+x^3+x^2+x+1\right)}{1+\left(x+1\right)+\left(x^2+x+1\right)+.....+\left(x^4+x^3+x^2+x+1\right)}\)
\(=\frac{1+2+....+6}{1+2+....+5}=\frac{\frac{6\left(5+1\right)}{2}}{\frac{5\left(5+1\right)}{2}}=\frac{7}{5}\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2020
\(a=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}{\left(x-1\right)\left(x^2+x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x+1\right)\left(x^2+1\right)}{x^2+x-1}=\frac{4}{1}=4\)
\(b=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{x^4-x^3+x^2-x+1}{x^2-x+1}=\frac{5}{3}\)
\(c=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)^2}{\left(x^2+1\right)\left(x^2-9\right)}=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x+3\right)}=\frac{0}{60}=0\)
\(d=\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{x^2-2x+1}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2x-2}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=10\)
\(e=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)
\(f=\lim\limits_{x\rightarrow-2}\frac{\left(x+2\right)\left(x-2\right)\left(x^2+4\right)}{\left(x+2\right)x^2}=\lim\limits_{x\rightarrow-2}\frac{\left(x-2\right)\left(x^2+4\right)}{x^2}=-8\)
Hai câu d, e khai triển thì dài quá nên làm biếng sử dụng L'Hopital
HT
24 tháng 1 2021
a/ \(=\lim\limits_{h\rightarrow0}\dfrac{2x^3+6x^2h+6xh^2+2h^3-2x^3}{h}\)
\(=\lim\limits_{h\rightarrow0}\dfrac{6xh^2+6x^2h+2h^3}{h}=\lim\limits_{h\rightarrow0}\left(6xh+6x^2+2h^2\right)=6x^2\)
b/ Xet day :\(S=x+x^2+....+x^{2021}\)
Day co \(\left\{{}\begin{matrix}u_1=x\\q=x\end{matrix}\right.\Rightarrow S=u_1.\dfrac{q^{2021}-1}{q-1}=x.\dfrac{x^{2021}-1}{x-1}\)
\(\Rightarrow\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^{2022}-x}{x-1}-2021}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{x^{2022}-x-2021x+2021}{\left(x-1\right)^2}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^{2022}}{x^2}-\dfrac{x}{x^2}-\dfrac{2021x}{x^2}+\dfrac{2021}{x^2}}{\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}+\dfrac{1}{x^2}}=\lim\limits_{x\rightarrow1}\dfrac{x^{2020}}{1}=1\)
HT
24 tháng 1 2021
Lam lai cau b, hinh nhu bi nham sang dang \(\dfrac{\infty}{\infty}\) roi
Xet day: \(S=x+x^2+...+x^{2021}\)
\(\Rightarrow S=x.\dfrac{x^{2021}-1}{x-1}=\dfrac{x^{2022}-x}{x-1}\)
\(\Rightarrow\lim\limits_{x\rightarrow1}\dfrac{x^{2022}-2022x+2021}{\left(x-1\right)^2}\)
L'Hospital: \(\Rightarrow...=\lim\limits_{x\rightarrow1}\dfrac{2022x^{2021}-2022}{2\left(x-1\right)}=\lim\limits_{x\rightarrow1}\dfrac{2022.2021.x^{2020}}{2}=2043231\)
Is that true :v?
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 4 2020
\(A=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x^3+x^2+x-2\right)}{\left(x-1\right)\left(x^2+x+3\right)}=\lim\limits_{x\rightarrow1}\frac{x^3+x^2+x-2}{x^2+x+3}=\frac{1}{5}\)
\(B=\lim\limits_{x\rightarrow3}\frac{-x^2+2x+3}{\left(x-1\right)\left(x-3\right)\left(\sqrt{2x+3}+x\right)}=\lim\limits_{x\rightarrow3}\frac{\left(3-x\right)\left(x+1\right)}{\left(x-1\right)\left(x-3\right)\left(\sqrt{2x+3}+x\right)}\)
\(=\lim\limits_{x\rightarrow3}\frac{-x-1}{\left(x-1\right)\left(\sqrt{2x+3}+x\right)}=\frac{-4}{2.\left(3+3\right)}=-\frac{1}{3}\)
TL
1
16 tháng 7 2016
mình ko ấn dấu lim, bn tự biết thêm vào nhé:
\(\frac{x^2-4x+3}{\sqrt{4x+5}-3}=\frac{\left(x-3\right)\left(x-1\right)\left(\sqrt{4x+5}+3\right)}{4x-4}=\frac{\left(x-3\right)\left(\sqrt{4x+5}+3\right)}{4}=-3\)
AH
Akai Haruma
Giáo viên
22 tháng 2 2023
Lời giải:
a. \(\lim\limits_{x\to 1+}(x^3+x+1)=3>0\)
\(\lim\limits_{x\to 1+}(x-1)=0\) và $x-1>0$ khi $x>1$
\(\Rightarrow \lim\limits_{x\to 1+}\frac{x^3+x+1}{x-1}=+\infty\)
b.
\(\lim\limits_{x\to -1+}(3x+2)=-1<0\)
\(\lim\limits_{x\to -1+}(x+1)=0\) và $x+1>0$ khi $x>-1$
\(\Rightarrow \lim\limits_{x\to -1+}\frac{3x+2}{x+1}=-\infty\)
c.
\(\lim\limits_{x\to 2-}(x-15)=-17<0\)
\(\lim\limits_{x\to 2-}(x-2)=0\) và $x-2<0$ khi $x<2$
\(\Rightarrow \lim\limits_{x\to 2-}\frac{x-15}{x-2}=+\infty\)
\(\lim\limits_{x\rightarrow1}\frac{x^4+x^3-2}{x^5-x^2}=\lim\limits_{x\rightarrow1}\frac{x^4-1+x^3-1}{x^2\left(x^3-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{\left(x^2-1\right)\left(x^2+1\right)+\left(x-1\right)\left(x^2+x+1\right)}{x^2\left(x-1\right)\left(x^2+x+1\right)}\)\(=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left[\left(x+1\right)\left(x^2+1\right)+\left(x^2+x+1\right)\right]}{x^2\left(x-1\right)\left(x^2+x+1\right)}\)\(=\lim\limits_{x\rightarrow1}\frac{\left[\left(x+1\right)\left(x^2+1\right)+\left(x^2+x+1\right)\right]}{x^2\left(x^2+x+1\right)}\)=\(\frac{7}{3}\)
=lim x^2(x^2+x) - 2 \ x^2(x^3-1)=lim(x^2+x)\(x^3-1)=lim 2\-2=-1