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.
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HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
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) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} - x - 2} \right) = {1^2} - 1 - 2 = - 2\\\mathop {\lim }\limits_{x \to 1} f\left( x \right) - \mathop {\lim }\limits_{x \to 1} g\left( x \right) = 0 - 2 = - 2\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} f\left( x \right) - \mathop {\lim }\limits_{x \to 1} g\left( x \right).\end{array}\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
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) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{x + 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to 1} \left( {x - 1} \right) = 1 - 1 = 0\\\frac{{\mathop {\lim }\limits_{x \to 1} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 1} g\left( x \right)}} = \frac{0}{2} = 0\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to 1} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 1} g\left( x \right)}}.\end{array}\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
a) \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} x = 1\)
b) \(f\left( 1 \right) = 1 \Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right).\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
\(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} x = 1\).
\(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( { - {x^2}} \right) = - {1^2} = - 1\).
Vì \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {1^ - }} {\rm{ }}f\left( x \right)\) nên không tồn tại \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\).
QT
Quoc Tran Anh Le
Giáo viên
22 tháng 9 2023
a) Giả sử \(\left( {{x_n}} \right)\) là dãy số bất kì, \({x_n} > - 1\) và \({x_n} \to - 1\). Khi đó \(f\left( {{x_n}} \right) = x_n^2 + 2\)
Ta có: \(\lim f\left( {{x_n}} \right) = \lim \left( {x_n^2 + 2} \right) = \lim x_n^2 + \lim 2 = {\left( { - 1} \right)^2} + 2 = 3\)
Vậy \(\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = 3\).
Giả sử \(\left( {{x_n}} \right)\) là dãy số bất kì, \({x_n} < - 1\) và \({x_n} \to - 1\). Khi đó \(f\left( {{x_n}} \right) = 1 - 2{x_n}\).
Ta có: \(\lim f\left( {{x_n}} \right) = \lim \left( {1 - 2{x_n}} \right) = \lim 1 - \lim \left( {2{x_n}} \right) = \lim 1 - 2\lim {x_n} = 1 - 2.\left( { - 1} \right) = 3\)
Vậy \(\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = 3\).
b) Vì \(\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ - }} {\rm{ }}f\left( x \right) = 3\) nên \(\mathop {\lim }\limits_{x \to - 1} f\left( x \right) = 3\).
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
a) \(\mathop {\lim }\limits_{x \to 2} \left[ {\left( {x + 1} \right)\left( {{x^2} + 2x} \right)} \right] = \mathop {\lim }\limits_{x \to 2} \left( {x + 1} \right).\mathop {\lim }\limits_{x \to 2} \left( {{x^2} + 2x} \right) = \left( {2 + 1} \right).\left( {{2^2} + 2.2} \right) = 24\)
b) \(\mathop {\lim }\limits_{x \to 2} \sqrt {{x^2} + x + 3} = \sqrt {\mathop {\lim }\limits_{x \to 2} \left( {{x^2} + x + 3} \right)} = \sqrt {\mathop {\lim }\limits_{x \to 2} {x^2} + \mathop {\lim }\limits_{x \to 2} x + \mathop {\lim }\limits_{x \to 2} 3} = \sqrt {{2^2} + 2 + 3} = 3\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 3 2021
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)\Rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=2\)
22 tháng 9 2023
a: \(\lim\limits_{x\rightarrow-2}x^2-7x+4=\left(-2\right)^2-7\cdot\left(-2\right)+4=22\)
b: \(\lim\limits_{x\rightarrow3}\dfrac{x-3}{x^2-9}=\lim\limits_{x\rightarrow3}\dfrac{1}{x+3}=\dfrac{1}{3+3}=\dfrac{1}{6}\)
c: \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-x-8}{3+\sqrt{x+8}}\cdot\dfrac{1}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{-1}{3+\sqrt{x+8}}\)
\(=-\dfrac{1}{6}\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
a) \(\mathop {\lim }\limits_{x \to - 1} \left( {3{x^2} - x + 2} \right) = \mathop {\lim }\limits_{x \to - 1} \left( {3{x^2}} \right) - \mathop {\lim }\limits_{x \to - 1} x + \mathop {\lim }\limits_{x \to - 1} 2\)
\( = 3\mathop {\lim }\limits_{x \to - 1} \left( {{x^2}} \right) - \mathop {\lim }\limits_{x \to - 1} x + \mathop {\lim }\limits_{x \to - 1} 2 = 3.{\left( { - 1} \right)^2} - \left( { - 1} \right) + 2 = 6\)
b) \(\mathop {\lim }\limits_{x \to 4} \frac{{{x^2} - 16}}{{x - 4}} = \mathop {\lim }\limits_{x \to 4} \frac{{\left( {x - 4} \right)\left( {x + 4} \right)}}{{x - 4}} = \mathop {\lim }\limits_{x \to 4} \left( {x + 4} \right) = \mathop {\lim }\limits_{x \to 4} x + \mathop {\lim }\limits_{x \to 4} 4 = 4 + 4 = 8\)
c) \(\mathop {\lim }\limits_{x \to 2} \frac{{3 - \sqrt {x + 7} }}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {3 - \sqrt {x + 7} } \right)\left( {3 + \sqrt {x + 7} } \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{{3^2} - \left( {x + 7} \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}}\)
\( = \mathop {\lim }\limits_{x \to 2} \frac{{2 - x}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{ - \left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{ - 1}}{{3 + \sqrt {x + 7} }}\)
\( = \frac{{\mathop {\lim }\limits_{x \to 2} \left( { - 1} \right)}}{{\mathop {\lim }\limits_{x \to 2} 3 + \sqrt {\mathop {\lim }\limits_{x \to 2} x + \mathop {\lim }\limits_{x \to 2} 7} }} = \frac{{ - 1}}{{3 + \sqrt {2 + 7} }} = - \frac{1}{6}\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
a) \(\mathop {\lim }\limits_{x \to 2} \left( {{x^2} - 4x + 3} \right) = \mathop {\lim }\limits_{x \to 2} {x^2} - \mathop {\lim }\limits_{x \to 2} \left( {4x} \right) + 3 = {2^2} - 4.2 + 3 = - 1\)
b) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 5x + 6}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x - 2} \right)}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \left( {x - 2} \right) = \mathop {\lim }\limits_{x \to 3} x - 2 = 3 - 2 = 1\)
c) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x + 1}} = \frac{1}{{\sqrt 1 + 1}} = \frac{1}{2}\)
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 \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\)