Quan sát đồ thị hàm số \(f\left( x \right) = x\) ở Hình 11.
a) Tính \(\mathop {\lim }\limits_{x \to 1} f\left( x \right).\)
b) So sánh \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\) với \(f\left( 1 \right).\)
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) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} - x - 2} \right) = {1^2} - 1 - 2 = - 2\\\mathop {\lim }\limits_{x \to 1} f\left( x \right) - \mathop {\lim }\limits_{x \to 1} g\left( x \right) = 0 - 2 = - 2\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} f\left( x \right) - \mathop {\lim }\limits_{x \to 1} g\left( x \right).\end{array}\)
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) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{x + 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to 1} \left( {x - 1} \right) = 1 - 1 = 0\\\frac{{\mathop {\lim }\limits_{x \to 1} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 1} g\left( x \right)}} = \frac{0}{2} = 0\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to 1} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 1} g\left( x \right)}}.\end{array}\)
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 \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\)
Cho hàm số \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{ - {x^2}}&{khi\,\,x < 1}\\x&{khi\,\,x \ge 1}\end{array}} \right.\).
Tìm các giới hạn \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right);\mathop {\lim }\limits_{x \to {1^ - }} {\rm{ }}f\left( x \right);\mathop {\lim }\limits_{x \to 1} f\left( x \right)\) (nếu có).
\(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} x = 1\).
\(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( { - {x^2}} \right) = - {1^2} = - 1\).
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 \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\).
Cho hàm số \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{1 - 2x}&{khi\,\,x \le - 1}\\{{x^2} + 2}&{khi\,\,x > - 1}\end{array}} \right.\).
Tìm các giới hạn \(\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right),\mathop {\lim }\limits_{x \to - {1^ - }} {\rm{ }}f\left( x \right)\) và \(\mathop {\lim }\limits_{x \to - 1} f\left( x \right)\) (nếu có).
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 \(\mathop {\lim }\limits_{x \to - 1} f\left( x \right) = 3\).
Tính:
a) \(\mathop {\lim }\limits_{x \to 2} \left[ {\left( {x + 1} \right)\left( {{x^2} + 2x} \right)} \right];\)
b) \(\mathop {\lim }\limits_{x \to 2} \sqrt {{x^2} + x + 3} .\)
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) \(\mathop {\lim }\limits_{x \to 2} \sqrt {{x^2} + x + 3} = \sqrt {\mathop {\lim }\limits_{x \to 2} \left( {{x^2} + x + 3} \right)} = \sqrt {\mathop {\lim }\limits_{x \to 2} {x^2} + \mathop {\lim }\limits_{x \to 2} x + \mathop {\lim }\limits_{x \to 2} 3} = \sqrt {{2^2} + 2 + 3} = 3\)
Cho hàm số \(y = f(x)\) xác định trên khoảng \((a;b)\) và \({x_0} \in (a;b)\). Điều kiện cần và đủ để hàm số \(y = f(x)\) liên tục tại \({x_0}\) là:
A. \(\mathop {\lim }\limits_{x \to x_0^ + } f(x) = f\left( {{x_0}} \right)\).
B. \(\mathop {\lim }\limits_{x \to x_0^ - } f(x) = f\left( {{x_0}} \right)\).
C. \(\mathop {\lim }\limits_{x \to x_0^ + } f(x) = \mathop {\lim }\limits_{x \to x_0^ - } f(x)\).
D. \(\mathop {\lim }\limits_{x \to x_0^ + } f(x) = \mathop {\lim }\limits_{x \to x_0^ - } f(x) = f\left( {{x_0}} \right)\).
Theo lí thuyết ta chọn đáp án D.
Cho hàm số \(y = f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{x + 1}&{khi\,\,1 < x \le 2}\\k&{khi\,\,x = 1}\end{array}} \right.\).
a) Xét tính liên tục của hàm số tại mỗi điểm \({x_0} \in \left( {1;2} \right)\).
b) Tìm \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right)\) và so sánh giá trị này với \(f\left( 2 \right)\).
c) Với giá trị nào của \(k\) thì \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = k\)?
a) Với mọi điểm \({x_0} \in \left( {1;2} \right)\), ta có: \(f\left( {{x_0}} \right) = {x_0} + 1\).
\(\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = \mathop {\lim }\limits_{x \to {x_0}} \left( {x + 1} \right) = {x_0} + 1\).
Vì \(\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = f\left( {{x_0}} \right) = {x_0} + 1\) nên hàm số \(y = f\left( x \right)\) liên tục tại mỗi điểm \({x_0} \in \left( {1;2} \right)\).
b) \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {x + 1} \right) = 2 + 1 = 3\).
\(f\left( 2 \right) = 2 + 1 = 3\).
\( \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = f\left( 2 \right)\).
c) \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {x + 1} \right) = 1 + 1 = 2\)
\(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = k \Leftrightarrow 2 = k \Leftrightarrow k = 2\)
Vậy với \(k = 2\) thì \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = k\).
Tính các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to 2} \left( {{x^2} - 4x + 3} \right);\)
b) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 5x + 6}}{{x - 3}};\)
c) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{x - 1}}.\)
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) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x + 1}} = \frac{1}{{\sqrt 1 + 1}} = \frac{1}{2}\)
Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to - {1^ + }} \frac{1}{{x + 1}}\);
b) \(\mathop {\lim }\limits_{x \to - \infty } \left( {1 - {x^2}} \right)\);
c) \(\mathop {\lim }\limits_{x \to {3^ - }} \frac{x}{{3 - x}}\).
a: \(\lim\limits_{x\rightarrow-1^+}x+1=0\)
=>\(\lim\limits_{x\rightarrow-1^+}\dfrac{1}{x+1}=+\infty\)
b: \(\lim\limits_{x\rightarrow-\infty}1-x^2=\lim\limits_{x\rightarrow-\infty}\left[x^2\left(\dfrac{1}{x^2}-1\right)\right]\)
\(=-\infty\)
c: \(\lim\limits_{x\rightarrow3^-}\dfrac{x}{3-x}=\lim\limits_{x\rightarrow3^-}=\dfrac{-x}{x-3}\)
\(\lim\limits_{x\rightarrow3^-}x-3=0\)
\(\lim\limits_{x\rightarrow3^-}-x=3>0\)
=>\(\lim\limits_{x\rightarrow3^-}\dfrac{x}{3-x}=+\infty\)