\(\lim\limits_{x\rightarrow1}\dfrac{-x^4-5x+6}{x^2+5x-6}\)
K
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TD
2
23 tháng 2 2021
\(\lim\limits_{x\rightarrow1}\dfrac{x-5x^5+4x^6}{\left(1-x\right)^2}=\lim\limits_{x\rightarrow1}\dfrac{x\left(1-5x^4+4x^5\right)}{\left(1-x\right)^2}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{x\left(4x^5-8x^4+4x^3+3x^4-6x^3+3x^2+2x^3-4x^2+2x+x^2-2x+1\right)}{\left(1-x\right)^2}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{x\left(x-1\right)^2\left(4x^3+3x^2+2x+1\right)}{\left(x-1\right)^2}\)
\(=\lim\limits_{x\rightarrow1}x\left(4x^3+3x^2+2x+1\right)=10\)
23 tháng 2 2021
bằng 0 nhé, lên youtube gõ thầy nguyễn công trí xem đi b ơi , tính nhanh tắc nghiệm thoi mà
HT
7 tháng 2 2021
1/ \(=\lim\limits_{x\rightarrow0}\dfrac{3\left(1+3x\right)^2.3+4.4\left(1-4x\right)^3}{1}=...\left(thay-x-vo\right)\)
2/ \(=\lim\limits_{x\rightarrow2}\dfrac{2.2.x-5}{3x^2-3}=\dfrac{4.2-5}{3.4-3}=\dfrac{1}{3}\)
3/ \(=\lim\limits_{x\rightarrow1}\dfrac{4x^3-3}{3x^2+2}=\dfrac{4.1-3}{3.1-2}=1\)
Xai L'Hospital nhe :v
23 tháng 2 2021
câu 1 bạn lm kiểu j vậy chả hiểu luôn bạn có thể lm lại chi tiết hơn dc ko
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 3 2022
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+2}+\sqrt{5x+4}-5}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+2}-2+\sqrt{5x+4}-3}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{2\left(x-1\right)}{\sqrt{2x+2}+2}+\dfrac{5\left(x-1\right)}{\sqrt{5x+4}+3}}{x-1}=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\sqrt{2x+2}+2}+\dfrac{5}{\sqrt{5x+4}+3}\right)=\dfrac{2}{2+2}+\dfrac{5}{3+3}=...\)
Đề câu b là \(...\sqrt{90-6x}\) hay \(\sqrt{9-6x}\) vậy em? Hình như cái sau mới có lý
C
1
RH
26 tháng 12 2023
\(\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{x^2-3x+2}}{x^2-5x+4}=\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{\left(1-x\right)\left(2-x\right)}}{\left(1-x\right)\left(4-x\right)}\\ =\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{2-x}}{\left(4-x\right)\sqrt{1-x}}\)
\(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow1^-}\sqrt{2-x}=1>0\\\lim\limits_{x\rightarrow1^-}\left(4-x\right)\sqrt{1-x}=0\\x< 1\rightarrow\left(4-x\right)\sqrt{1-x}>0\end{matrix}\right.\\ \rightarrow\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{x^2-3x+2}}{x^2-5x+4}=\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{2-x}}{\left(4-x\right)\sqrt{1-x}}=+\infty\)
C
1
25 tháng 4 2017
a/ \(\lim\limits_{x\to 1} f(x)=\frac{x^{2}-5x + 6}{x-2} \)
\(<=>\lim\limits_{x\to 1} f(x)=\dfrac{(x-3)(x-2)}{x-2} \)
<=>\(\lim\limits_{x\to 1} f(x)=x-3 \)
\(<=>\lim\limits_{x\to 1} f(x)=-2\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 2 2021
\(a=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-1+1-\sqrt[3]{2x+1}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x}{\sqrt[]{4x+1}+1}+\dfrac{-2x}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{4x+1}+1}+\dfrac{-2}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}\right)=...\)
\(b=\lim\limits_{x\rightarrow1}\dfrac{4\left(x-1\right)\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(x-1\right)\left(\sqrt[]{4x+5}+3\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{4\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(\sqrt[]{4x+5}+3\right)}=...\)
\(c=\lim\limits_{x\rightarrow-1}\dfrac{\left(2x+3\right)^{\dfrac{1}{4}}+\left(2+3x\right)^{\dfrac{1}{3}}}{\left(x+2\right)^{\dfrac{1}{2}}-1}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{1}{2}\left(2x+3\right)^{-\dfrac{3}{4}}+\left(2+3x\right)^{-\dfrac{2}{3}}}{\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}=3\)
10 tháng 11 2023
a: \(\lim\limits_{x\rightarrow3}\dfrac{x^2-9}{x^2-5x+6}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{x+3}{x-2}=\dfrac{3+3}{3-2}=\dfrac{6}{1}=6\)
b: \(\lim\limits_{x\rightarrow5}\dfrac{x^2-5x}{x-5}=\lim\limits_{x\rightarrow5}\dfrac{x\left(x-5\right)}{x-5}=\lim\limits_{x\rightarrow5}x=5\)
c: \(\lim\limits_{x\rightarrow-3}\dfrac{x^2-3x}{2x^2+9x+9}\)
\(=\lim\limits_{x\rightarrow-3}\dfrac{x\left(x-3\right)}{2x^2+6x+3x+9}\)
\(=\lim\limits_{x\rightarrow-3}\dfrac{\left(-3\right)\left(-3-3\right)}{\left(-3+3\right)\left(2\cdot\left(-3\right)+3\right)}\)
\(=\lim\limits_{x\rightarrow-3}\dfrac{18}{0\cdot\left(-3\right)}=-\infty\)
\(\lim\limits_{x\rightarrow1}\dfrac{-x^4-5x+6}{x^2+5x-6}=\lim\limits_{x\rightarrow1}\dfrac{-\left(x-1\right)\left(x^3+x^2+x+6\right)}{\left(x-1\right)\left(x+6\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{-\left(x^3+x^2+x+6\right)}{x+6}=\dfrac{-9}{7}\)