Tìm các số thực a, b thoả mãn:
\(\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left[\left(a^3+b^3\right)x^2-\left(x+a^2b\right)\sqrt{x^2+2\left(ab\right)^2}\right]}{x-b-1}\)
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TL
7 tháng 11 2023
Em là tám lại ạ
Em là duy khôi ạ
Em là văn tam ạ
Em là mạnh Tuấn ạ
8 tháng 11 2023
a: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x-2}-\dfrac{12}{x^3-8}\right)\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x+4-12}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x-8}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x+4}{x^2+2x+4}\)
\(=\dfrac{2+4}{2^2+2\cdot2+4}=\dfrac{6}{4+4+4}=\dfrac{6}{12}=\dfrac{1}{2}\)
b: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\)
\(=\lim\limits_{x\rightarrow2}\left(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}\right)\)
\(=\lim\limits_{x\rightarrow2}\left(\dfrac{x-3+x-1}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\right)\)
\(=\lim\limits_{x\rightarrow2}\dfrac{2x-4}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{2}{\left(x-3\right)\left(x-1\right)}=\dfrac{2}{\left(2-3\right)\left(2-1\right)}=-2\)
d: \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-\sqrt[3]{x^3-1}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-x+x-\sqrt[3]{x^3-1}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}+\dfrac{x^3-x^3+1}{\sqrt[3]{x^2}+x\cdot\sqrt[3]{x^3-1}+\sqrt[3]{\left(x^3-1\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{1}{\sqrt{x^2+1}+x}+\dfrac{1}{\sqrt[3]{x^2}+x\cdot\sqrt[3]{x^3-1}+\sqrt[3]{\left(x^3-1\right)^2}}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\dfrac{1}{x}}{\sqrt{1+\dfrac{1}{x^2}}+1}+\dfrac{\dfrac{1}{x^2}}{\sqrt[3]{\dfrac{1}{x^4}}+\sqrt[3]{1-\dfrac{1}{x^3}}+\sqrt[3]{\left(1-\dfrac{1}{x^3}\right)^2}}\right)\)
=0
c: \(\lim\limits_{x\rightarrow+\infty}\left[x\cdot\left(\sqrt{x^2+1}-x\right)\right]\)
\(=\lim\limits_{x\rightarrow+\infty}\left[x\cdot\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}\right]\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x}{\sqrt{x^2+1}+x}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}+1}=\dfrac{1}{1+1}=\dfrac{1}{2}\)
e: \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{x^2+1-1}{\sqrt{x^2+1}+1}:\dfrac{x^2+16-16}{\sqrt{x^2+16}+4}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+16}+4}{\sqrt{x^2+1}+1}=\dfrac{4+4}{1+1}=\dfrac{8}{2}=4\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2022
\(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)+1}{x-2}\) hữu hạn \(\Rightarrow f\left(x\right)+1=0\) có nghiệm \(x=2\Rightarrow f\left(2\right)=-1\)
\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{f\left(x\right)+2x+1}-x}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{1}{\sqrt{f\left(x\right)+2x+1}+x}.\dfrac{\left(\sqrt{f\left(x\right)+2x+1}-x\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}.\dfrac{f\left(x\right)+1-x\left(x-2\right)}{x-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}.\left(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)+1}{x-2}-\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)}{x-2}\right)\)
\(=\dfrac{1}{4\left(\sqrt{4}+2\right)}.\left(a-2\right)=\dfrac{a-2}{16}\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 1
Em kiểm tra lại đề, chỗ \(f\left(x\right)-32\) kia có vẻ sai, vì như thế thì biểu thức đã cho ko phải dạng vô định
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 3 2022
Do \(\lim\limits_{x\rightarrow3}\dfrac{f\left(x\right)-2}{x-3}\) hữu hạn \(\Rightarrow f\left(x\right)-2=0\) có nghiệm \(x=3\)
Hay \(f\left(3\right)-2=0\Rightarrow f\left(3\right)=2\)
\(\Rightarrow I=\lim\limits_{x\rightarrow3}\left(\dfrac{f\left(x\right)-2}{x-3}\right).\dfrac{1}{\sqrt{5f\left(x\right)+6}+1}=\dfrac{1}{4}.\dfrac{1}{\sqrt{5.f\left(3\right)+6}+1}\)
\(=\dfrac{1}{4}.\dfrac{1}{\sqrt{5.2+6}+1}=\dfrac{1}{20}\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 3 2022
Do \(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)-3}{x-2}=5\Rightarrow\) chọn \(f\left(x\right)=5\left(x-2\right)+3=5x-7\)
\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt[]{5x-7+6}-\sqrt[3]{x+25}}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\sqrt[]{5x-1}-3+3-\sqrt[3]{x+25}}{x-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{5\left(x-2\right)}{\sqrt[]{5x-1}+3}-\dfrac{x-2}{9+3\sqrt[3]{x+25}+\sqrt[3]{\left(x+25\right)^2}}}{x-2}\)
\(=\lim\limits_{x\rightarrow2}\left(\dfrac{5}{\sqrt[]{5x-1}+3}-\dfrac{1}{9+3\sqrt[3]{x+25}+\sqrt[3]{\left(x+25\right)^2}}\right)=\dfrac{5}{3+3}-\dfrac{1}{9+9+9}=\dfrac{43}{54}\)
V
1
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 3 2022
Giới hạn đã cho hữu hạn khi và chỉ khi \(b=1\)
Khi đó:
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2-ax+1}-x\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{-ax+1}{\sqrt{x^2-ax+1}+x}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{-a+\dfrac{1}{x}}{\sqrt{1-\dfrac{a}{x}+\dfrac{1}{x^2}}+1}=-\dfrac{a}{2}\)
\(\Rightarrow-\dfrac{a}{2}=2\Rightarrow a=-4\)
Vậy \(\left(a;b\right)=\left(-4;1\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2020
\(a=\lim\limits_{x\rightarrow-\infty}\left(\frac{-x^2}{\sqrt[3]{\left(x^3-x^2\right)^2}+x\sqrt[3]{x^3-x^2}+x^2}\right)=\lim\limits_{x\rightarrow-\infty}\left(\frac{-1}{\sqrt[3]{\left(1-\frac{1}{x}\right)^3}+\sqrt[3]{1-\frac{1}{x}}+1}\right)=-\frac{1}{3}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{5x^2-8x}{\sqrt[3]{\left(x^3+5x^2\right)^2}+\sqrt[3]{\left(x^3+5x^2\right)\left(x^3+8x\right)}+\sqrt[3]{\left(x^3+8x\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\frac{5-\frac{8}{x}}{\sqrt[3]{\left(1+\frac{5}{x}\right)^2}+\sqrt[3]{\left(1+\frac{5}{x}\right)\left(1+\frac{8}{x^2}\right)}+\sqrt[3]{\left(1+\frac{8}{x^2}\right)^2}}=\frac{5}{3}\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{1}{\sqrt[3]{\left(x^3+1\right)^2}+x\sqrt[3]{x^3+1}+x^2}=\frac{1}{+\infty}=0\)
Bài 2:
\(a=\lim\limits_{x\rightarrow1^-}\left(\frac{1-x}{\left(x-1\right)\left(x+1\right)}\right)=\lim\limits_{x\rightarrow1^-}\frac{-1}{x+1}=-\frac{1}{2}\)
\(b=\lim\limits_{x\rightarrow1^+}\left(\frac{x^2+x+1-3}{\left(1-x\right)\left(x^2+x+1\right)}\right)=\lim\limits_{x\rightarrow1^+}\frac{\left(x-1\right)\left(x+2\right)}{\left(1-x\right)\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow1^+}\frac{-x-2}{x^2+x+1}=-1\)
\(c=\lim\limits_{x\rightarrow2^+}\left(\frac{1}{\left(x-1\right)\left(x-2\right)}-\frac{1}{\left(x-2\right)\left(x-3\right)}\right)=\lim\limits_{x\rightarrow2^+}\frac{-2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
Do \(x\rightarrow2^+\Rightarrow x>2\Rightarrow x-2>0\Rightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\rightarrow0^-\)
\(\Rightarrow\lim\limits_{x\rightarrow2^+}\frac{-2}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=+\infty\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 1 2021
Do \(\lim\limits_{x\rightarrow-1}\dfrac{2f\left(x\right)+1}{x+1}=5\) hữu hạn nên \(2f\left(x\right)+1=0\) phải có nghiệm \(x=-1\)
\(\Leftrightarrow2f\left(-1\right)=-1\Leftrightarrow f\left(-1\right)=-\dfrac{1}{2}\)
Đoạn dưới tự hiểu là \(\lim\limits_{x\rightarrow-1}\) (vì kí tự lim rất rắc rối)
\(I=\dfrac{\left[4f\left(x\right)+3\right]\left[\sqrt{4f^2\left(x\right)+2f\left(x\right)+4}-2\right]+2\left[4f\left(x\right)+3\right]-2}{x^2-1}\)
\(=\dfrac{\left[4f\left(x\right)+3\right]\left[4f^2\left(x\right)+2f\left(x\right)\right]}{\left(x+1\right)\left(x-1\right)\left[\sqrt{4f^2\left(x\right)+2f\left(x\right)+4}+2\right]}+\dfrac{4\left[2f\left(x\right)+1\right]}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{2f\left(x\right)+1}{x+1}.\dfrac{f\left(x\right).\left[4f\left(x\right)+3\right]}{x-1}+\dfrac{2f\left(x\right)+1}{x+1}.\dfrac{4}{x-1}\)
\(=5.\dfrac{f\left(-1\right).\left[4f\left(-1\right)+3\right]}{-2}+5.\dfrac{4}{-2}=\dfrac{5.\left(-\dfrac{1}{2}\right)\left(-2+3\right)}{-2}+5.\dfrac{4}{-2}=...\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 1 2021
Không phải dạng, nó chỉ là ứng dụng kiến thức cơ bản về giới hạn của hàm thôi
