Biết lim x -> +∞ f(x) = M ;lim x -> +∞ g(x) = 0 Chọn khẳng định đúng? A. Lim x -> +∞ f(x)/g(x)= +∞ B. Lim x -> +∞ = f(x)/g(x)= -∞ C. Lim x -> +∞ f(x)/g(x)=0 D. Limx -> +∞ [g(x).f(x)]=0
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Biết rằng hàm số \(f\left( x \right)\) thỏa mãn \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = 3\) và \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = 5.\) Trong trường hợp này có tồn tại giới hạn \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\) hay không? Giải thích.
Vì \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = 3 \ne \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = 5\) nên không tồn tại giới hạn \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\)
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Biết \(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)-3}{x-2}=5\). Tính \(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{f\left(x\right)+6}-\sqrt[3]{x+25}}{x-2}\)
Do \(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)-3}{x-2}=5\Rightarrow\) chọn \(f\left(x\right)=5\left(x-2\right)+3=5x-7\)
\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt[]{5x-7+6}-\sqrt[3]{x+25}}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\sqrt[]{5x-1}-3+3-\sqrt[3]{x+25}}{x-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{5\left(x-2\right)}{\sqrt[]{5x-1}+3}-\dfrac{x-2}{9+3\sqrt[3]{x+25}+\sqrt[3]{\left(x+25\right)^2}}}{x-2}\)
\(=\lim\limits_{x\rightarrow2}\left(\dfrac{5}{\sqrt[]{5x-1}+3}-\dfrac{1}{9+3\sqrt[3]{x+25}+\sqrt[3]{\left(x+25\right)^2}}\right)=\dfrac{5}{3+3}-\dfrac{1}{9+9+9}=\dfrac{43}{54}\)
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biết lim x->1 =[ 2f(x)-2f(1) ]/ (x-1) =5 , tìm f(1) hoặc f(x)
8 tháng 3 2021 lúc 15:50
cho hàm số f(x) thỏa mãn: \(\lim\limits_{x\rightarrow1^+}f\left(x\right)=2\) và \(\lim\limits_{x\rightarrow1^-}f\left(x\right)=2\). tính giá trị \(\lim\limits_{x\rightarrow1}f\left(x\right)=?\)
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)\Rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=2\)
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cho \(f\left(x\right)=\left\{{}\begin{matrix}x^2-3\\x+3\end{matrix}\right.\) \(x\ge3\);\(x< 3\)
a) tính \(\lim\limits_{x\rightarrow3^+}f\left(x\right)=?\)
\(\lim\limits_{x\rightarrow3^-}f\left(x\right)=?\)
b) tính \(\lim\limits_{x\rightarrow3}f\left(x\right)\) nếu có
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
=>\(\lim\limits_{x\rightarrow3}f\left(x\right)=6\)
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nếu lim f(x)=L>0, lim g(x)=-vô cùng thì kết quả của giới hạn lim f(x).g(x) là:
A/ - vô cùng
B/ 0
C/ + vô cùng
D/ L
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Cho hai hàm số fleft( x right) {x^2} - 1,gleft( x right) x + 1.a) Tính mathop {lim }limits_{x to 1} fleft( x right) và mathop {lim }limits_{x to 1} gleft( x right).b) Tính mathop {lim }limits_{x to 1} left[ {fleft( x right) + gleft( x right)} right]và so sánh mathop {lim }limits_{x to 1} fleft( x right) + mathop {lim }limits_{x to 1} gleft( x right).c) Tính mathop {lim }limits_{x to 1} left[ {fleft( x right) - gleft( x right)} right]và so sánh mathop {lim }limits_{x to 1} fleft( x right) - mat...
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Cho hai hàm số \(f\left( x \right) = {x^2} - 1,g\left( x \right) = x + 1.\)
a) Tính \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\) và \(\mathop {\lim }\limits_{x \to 1} g\left( x \right).\)
b) Tính \(\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) + g\left( x \right)} \right]\)và so sánh \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) + \mathop {\lim }\limits_{x \to 1} g\left( x \right).\)
c) Tính \(\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) - g\left( x \right)} \right]\)và so sánh \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) - \mathop {\lim }\limits_{x \to 1} g\left( x \right).\)
d) Tính \(\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right).g\left( x \right)} \right]\)và so sánh \(\mathop {\lim }\limits_{x \to 1} f\left( x \right).\mathop {\lim }\limits_{x \to 1} g\left( x \right).\)
e) Tính \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{g\left( x \right)}}\)và so sánh \(\frac{{\mathop {\lim }\limits_{x \to 1} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 1} g\left( x \right)}}.\)
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) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} - x - 2} \right) = {1^2} - 1 - 2 = - 2\\\mathop {\lim }\limits_{x \to 1} f\left( x \right) - \mathop {\lim }\limits_{x \to 1} g\left( x \right) = 0 - 2 = - 2\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} f\left( x \right) - \mathop {\lim }\limits_{x \to 1} g\left( x \right).\end{array}\)
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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) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{x + 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to 1} \left( {x - 1} \right) = 1 - 1 = 0\\\frac{{\mathop {\lim }\limits_{x \to 1} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 1} g\left( x \right)}} = \frac{0}{2} = 0\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to 1} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 1} g\left( x \right)}}.\end{array}\)
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cho lim \(\dfrac{f\left(x\right)-5}{x-1}=4\) khi x->1 , lim \(\dfrac{g\left(x\right)-1}{x-1}=5\) khi x->1
tinh lim \(\dfrac{\sqrt{f\left(x\right)\times g\left(x\right)+4}-1}{x-1}\)khi x->1
Bạn tham khảo:
Nếu \(lim\) (x->1) \(\dfrac{f\left(x\right)-5}{x-1}=2\) và lim (x->1) \(\dfrac{g\left(x\right)-1}{x-1}=3\) thì lim (x->1... - Hoc24
Không giống hoàn toàn, nhưng cách làm thì giống hoàn toàn
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27 tháng 1 2021 lúc 19:06
Nếu \(lim\) (x->1) \(\dfrac{f\left(x\right)-5}{x-1}=2\) và lim (x->1) \(\dfrac{g\left(x\right)-1}{x-1}=3\) thì lim (x->1) \(\dfrac{\sqrt{f\left(x\right).g\left(x\right)+4}-3}{x-1}\) bằng mấy
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: \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{f\left(x\right).g\left(x\right)+4}-3}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right).g\left(x\right)-5}{\left(x-1\right)\left(\sqrt{f\left(x\right).g\left(x\right)+4}+3\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left[f\left(x\right)-5\right].g\left(x\right)+5\left[g\left(x\right)-1\right]}{\left(x-1\right)\left(\sqrt{f\left(x\right).g\left(x\right)+4}+3\right)}\)
\(=\left(2.1+5.3\right).\dfrac{1}{\sqrt{5.1+4}+3}=\dfrac{17}{6}\)
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