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Do \(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)-3}{x-2}=5\Rightarrow\) chọn \(f\left(x\right)=5\left(x-2\right)+3=5x-7\)
\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt[]{5x-7+6}-\sqrt[3]{x+25}}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\sqrt[]{5x-1}-3+3-\sqrt[3]{x+25}}{x-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{5\left(x-2\right)}{\sqrt[]{5x-1}+3}-\dfrac{x-2}{9+3\sqrt[3]{x+25}+\sqrt[3]{\left(x+25\right)^2}}}{x-2}\)
\(=\lim\limits_{x\rightarrow2}\left(\dfrac{5}{\sqrt[]{5x-1}+3}-\dfrac{1}{9+3\sqrt[3]{x+25}+\sqrt[3]{\left(x+25\right)^2}}\right)=\dfrac{5}{3+3}-\dfrac{1}{9+9+9}=\dfrac{43}{54}\)
Mấy câu này bạn cần giải theo kiểu trắc nghiệm hay tự luận nhỉ?
a) \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} - 1} \right) = \mathop {\lim }\limits_{x \to 1} {x^2} - \mathop {\lim }\limits_{x \to 1} 1 = {1^2} - 1 = 0\)
\(\mathop {\lim }\limits_{x \to 1} g\left( x \right) = \mathop {\lim }\limits_{x \to 1} \left( {x + 1} \right) = \mathop {\lim }\limits_{x \to 1} x + \mathop {\lim }\limits_{x \to 1} 1 = 1 + 1 = 2\)
b) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) + g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} + x} \right) = {1^2} + 1 = 2\\\mathop {\lim }\limits_{x \to 1} f\left( x \right) + \mathop {\lim }\limits_{x \to 1} g\left( x \right) = 0 + 2 = 2\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) + g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} f\left( x \right) + \mathop {\lim }\limits_{x \to 1} g\left( x \right).\end{array}\)
c) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} - x - 2} \right) = {1^2} - 1 - 2 = - 2\\\mathop {\lim }\limits_{x \to 1} f\left( x \right) - \mathop {\lim }\limits_{x \to 1} g\left( x \right) = 0 - 2 = - 2\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} f\left( x \right) - \mathop {\lim }\limits_{x \to 1} g\left( x \right).\end{array}\)
d) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right).g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} \left[ {\left( {{x^2} - 1} \right)\left( {x + 1} \right)} \right] = \mathop {\lim }\limits_{x \to 1} \left( {{x^3} + {x^2} - x - 1} \right) = {1^3} + {1^2} - 1 - 1 = 0\\\mathop {\lim }\limits_{x \to 1} f\left( x \right).\mathop {\lim }\limits_{x \to 1} g\left( x \right) = 0.2 = 0\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right).g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} f\left( x \right).\mathop {\lim }\limits_{x \to 1} g\left( x \right).\end{array}\)
e) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{x + 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to 1} \left( {x - 1} \right) = 1 - 1 = 0\\\frac{{\mathop {\lim }\limits_{x \to 1} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 1} g\left( x \right)}} = \frac{0}{2} = 0\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to 1} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 1} g\left( x \right)}}.\end{array}\)
Chọn F(x)=5x-23
\(\lim\limits_{x\rightarrow5}\dfrac{f\left(x\right)-2}{x-5}=\lim\limits_{x\rightarrow5}\dfrac{5x-23-2}{x-5}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{5x-25}{x-5}=\lim\limits_{x\rightarrow5}\dfrac{5\left(x-5\right)}{x-5}=5\)
=>f(x)=5x-23 thỏa mãn yêu cầu đề bài
\(\lim\limits_{x\rightarrow5}\dfrac{\sqrt{3\cdot f\left(x\right)+10}+\sqrt{f^3\left(x\right)+1}-7}{x^2-25}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\sqrt{3\left(5x-23\right)+10}+\sqrt{\left(5x-23\right)^3+1}-7}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\sqrt{15x-59}+\sqrt{\left(5x-23\right)^3+1}-7}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\sqrt{15x-59}-4+\sqrt{\left(5x-23\right)^3+1}-3}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{15x-59-16}{\sqrt{15x-59}+4}+\dfrac{\left(5x-23\right)^3+1-9}{\sqrt{\left(5x-23\right)^3+1}+3}}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{15\left(x-5\right)}{\sqrt{15x-59}+4}+\dfrac{\left(5x-23\right)^3-8}{\sqrt{\left(5x-23\right)^3+1}+3}}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{15\left(x-5\right)}{\sqrt{15x-59}+4}+\dfrac{\left(5x-23-2\right)\left[\left(5x-23\right)^2+2\left(5x-23\right)+4\right]}{\sqrt{\left(5x-23\right)^3+1}+3}}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{15}{\sqrt{15x-59}+4}+\dfrac{5\cdot\left(25x^2-230x+529+10x-46+4\right)}{\sqrt{\left(5x-23\right)^3+1}+3}}{x+5}\)
\(=\dfrac{\dfrac{15}{\sqrt{15\cdot5-59}+4}+\dfrac{5\left(25\cdot5^2-220\cdot5+487\right)}{\sqrt{\left(5\cdot5-23\right)^3+1}+3}}{5+5}\)
\(=\dfrac{\dfrac{15}{8}+\dfrac{5\cdot12}{6}}{10}=\dfrac{19}{16}\)
Do \(\lim\limits_{x\rightarrow5}\dfrac{f\left(x\right)-2}{x-5}\) hữu hạn nên \(f\left(x\right)-2=0\) có nghiệm \(x=5\)
\(\Rightarrow f\left(5\right)=2\)
\(\lim\limits_{x\rightarrow5}\dfrac{\sqrt{3f\left(x\right)+10}-4+\sqrt{f^3\left(x\right)+1}-3}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{3\left[f\left(x\right)-2\right]}{\sqrt{3f\left(x\right)+10}+4}+\dfrac{\left[f\left(x\right)-2\right]\left[f^2\left(x\right)+2f\left(x\right)+4\right]}{\sqrt{f^3\left(x\right)+1}+3}}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{f\left(x\right)-2}{x-5}.\dfrac{3}{\sqrt{3f\left(x\right)+10}+4}+\dfrac{f\left(x\right)-2}{x-5}.\dfrac{f^2\left(x\right)+2f\left(x\right)+4}{\sqrt{f^3\left(x\right)+1}+3}}{x+5}\)
\(=\dfrac{5.\dfrac{3}{\sqrt{3.2+10}+4}+5.\dfrac{2^2+2.2+4}{\sqrt{2^3+1}+3}}{5+5}=\)
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 \(f\left(3\right)-2=0\Rightarrow f\left(3\right)=2\)
\(\Rightarrow I=\lim\limits_{x\rightarrow3}\left(\dfrac{f\left(x\right)-2}{x-3}\right).\dfrac{1}{\sqrt{5f\left(x\right)+6}+1}=\dfrac{1}{4}.\dfrac{1}{\sqrt{5.f\left(3\right)+6}+1}\)
\(=\dfrac{1}{4}.\dfrac{1}{\sqrt{5.2+6}+1}=\dfrac{1}{20}\)
\(\lim\limits_{x\rightarrow3}\dfrac{f\left(x\right)-80}{x-3}\) hữu hạn \(\Rightarrow f\left(3\right)=80\)
Sử dụng hẳng đẳng thức: \(a-b=\dfrac{a^4-b^4}{\left(a+b\right)\left(a^2+b^2\right)}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{\dfrac{f\left(x\right)-80}{\left[\sqrt[4]{f\left(x\right)+1}+3\right]\left[\sqrt[]{f\left(x\right)+1}+9\right]}}{\left(x-3\right)\left(2x-5\right)}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{f\left(x\right)-80}{x-3}.\dfrac{1}{\left[\sqrt[4]{f\left(x\right)+1}+3\right]\left[\sqrt[]{f\left(x\right)+1}+9\right]\left(2x-5\right)}\)
\(=5.\dfrac{1}{\left(\sqrt[4]{80+1}+3\right)\left(\sqrt[]{80+1}+9\right)\left(2.3-5\right)}\)
a: \(\lim\limits_{x\rightarrow3^+}f\left(x\right)=\lim\limits_{x\rightarrow3^+}x^2-3=3^2-3=6\)
\(\lim\limits_{x\rightarrow3^-}f\left(x\right)=\lim\limits_{x\rightarrow3^-}x+3=3+3=6\)
b: Vì \(\lim\limits_{x\rightarrow3^+}f\left(x\right)=\lim\limits_{x\rightarrow3^-}f\left(x\right)=6\)
nên hàm số tồn tại lim khi x=3
=>\(\lim\limits_{x\rightarrow3}f\left(x\right)=6\)
a.
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[3]{x^3+4x^2}-x\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{4x^2}{\sqrt[3]{\left(x^3+4x^2\right)^2}+x\sqrt[3]{x^3+4x^2}+x^2}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{4}{\sqrt[3]{\left(1+\dfrac{4}{x}\right)^2}+\sqrt[3]{1+\dfrac{4}{x}}+1}=\dfrac{4}{1+1+1}=\dfrac{4}{3}\)
b.
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{4x-1}{x-1}=\dfrac{3}{0}=+\infty\)
\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(7x+1\right)=8\)
Do \(x-1\rightarrow0\) khi \(x\rightarrow1\) nên \(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-5}{x-1}=2\) hữu hạn khi và chỉ khi \(f\left(x\right)-5=0\) có nghiệm \(x=1\)
\(\Leftrightarrow f\left(1\right)-5=0\Rightarrow f\left(1\right)=5\)
Tương tự ta có \(g\left(1\right)=1\)
Do đó: \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{f\left(x\right).g\left(x\right)+4}-3}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right).g\left(x\right)-5}{\left(x-1\right)\left(\sqrt{f\left(x\right).g\left(x\right)+4}+3\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left[f\left(x\right)-5\right].g\left(x\right)+5\left[g\left(x\right)-1\right]}{\left(x-1\right)\left(\sqrt{f\left(x\right).g\left(x\right)+4}+3\right)}\)
\(=\left(2.1+5.3\right).\dfrac{1}{\sqrt{5.1+4}+3}=\dfrac{17}{6}\)
Em làm như này được ko anh?
Tìm lim f(x) theo lim của x, rồi thế vô biểu thức, ví dụ như: \(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-5}{x-1}=2\Rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=\lim\limits_{x\rightarrow1}\left[2\left(x-1\right)+5\right]\)
Vậy là mình có thể chuyển từ tìm lim f(x) sang lim của hàm số chứa x
\(\lim\limits_{x\rightarrow+\infty}\dfrac{17}{x^2+1}=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{17}{x^2}}{1+\dfrac{1}{x^2}}=\dfrac{0}{1}=0\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{-2x^2+x-1}{3+x}=\lim\limits_{x\rightarrow+\infty}x\left(\dfrac{-2+\dfrac{1}{x}-\dfrac{1}{x^2}}{\dfrac{3}{x}+1}\right)\)
Do \(\lim\limits_{x\rightarrow+\infty}x=+\infty\)
\(\lim\limits_{x\rightarrow+\infty}\left(\dfrac{-2+\dfrac{1}{x}-\dfrac{1}{x^2}}{\dfrac{3}{x}+1}\right)=-2< 0\)
\(\Rightarrow\lim\limits_{x\rightarrow+\infty}x\left(\dfrac{-2+\dfrac{1}{x}-\dfrac{1}{x^2}}{\dfrac{3}{x}+1}\right)=-\infty\)
- Ta có:
Chọn A.