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\(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} x = 1\).
\(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( { - {x^2}} \right) = - {1^2} = - 1\).
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 \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\).
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\mathop {\lim }\limits_{x \to - 1} f\left( x \right) = 3\).
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
Mấy câu này bạn cần giải theo kiểu trắc nghiệm hay tự luận nhỉ?
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x=5\)
\(\Rightarrow f\left(5\right)=2\)
\(\lim\limits_{x\rightarrow5}\dfrac{\sqrt{3f\left(x\right)+10}-4+\sqrt{f^3\left(x\right)+1}-3}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{3\left[f\left(x\right)-2\right]}{\sqrt{3f\left(x\right)+10}+4}+\dfrac{\left[f\left(x\right)-2\right]\left[f^2\left(x\right)+2f\left(x\right)+4\right]}{\sqrt{f^3\left(x\right)+1}+3}}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{f\left(x\right)-2}{x-5}.\dfrac{3}{\sqrt{3f\left(x\right)+10}+4}+\dfrac{f\left(x\right)-2}{x-5}.\dfrac{f^2\left(x\right)+2f\left(x\right)+4}{\sqrt{f^3\left(x\right)+1}+3}}{x+5}\)
\(=\dfrac{5.\dfrac{3}{\sqrt{3.2+10}+4}+5.\dfrac{2^2+2.2+4}{\sqrt{2^3+1}+3}}{5+5}=\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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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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 \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. Áp dụng quy tắc L'Hopital
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+1}-1}{f\left(0\right)-f\left(x\right)}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{2\sqrt{x+1}}}{-f'\left(0\right)}=-\dfrac{1}{6}\)
2.
\(g'\left(x\right)=2x.f'\left(\sqrt{x^2+4}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\f'\left(\sqrt{x^2+4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x^2+4}=1\\\sqrt{x^2+4}=-2\end{matrix}\right.\)
2 pt cuối đều vô nghiệm nên \(g'\left(x\right)=0\) có đúng 1 nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
Chọn \(f\left(x\right)=5x+5\)
Khi đó: \(\lim\limits_{x\rightarrow1}\dfrac{5x-5}{\left(\sqrt{x}-1\right)\left(\sqrt{20x+29}+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{5\left(\sqrt{x}+1\right)}{\sqrt{20x+29}+3}=\dfrac{10}{7+3}=1\)
Chọn A
Cách 2: