\(L=\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{\tan^3x-3\tan x}{\cos\left(x+\frac{\pi}{6}\right)}=\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{\tan x\left(\tan^2x-3\right)}{\cos\left(x+\frac{\pi}{6}\right)}\)

    \(=\sqrt{3}\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{\left(\tan x-\sqrt{3}\right)\left(\tan x+\sqrt{3}\right)}{\sin\left(\frac{\pi}{3}-x\right)}=\sqrt{3}.2\sqrt{3}\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{\tan x-\sqrt{3}}{\sin\left(\frac{\pi}{3}-x\right)}\)

    \(=6\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{\sin\left(\frac{\pi}{3}-x\right)}{\cos x.\cos\frac{\pi}{3}\sin\left(\frac{\pi}{3}-x\right)}=-12\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{1}{\cos x}=-24\)

Tất cả đều ko phải dạng vô định, bạn cứ thay số vào tính thôi:

\(a=\frac{sin\left(\frac{\pi}{4}\right)}{\frac{\pi}{2}}=\frac{\sqrt{2}}{\pi}\)

\(b=\frac{\sqrt[3]{3.4-4}-\sqrt{6-2}}{3}=\frac{0}{3}=0\)

\(c=0.sin\frac{1}{2}=0\)

Bài 1:

\(a=\lim\limits_{x\rightarrow+\infty}\frac{\frac{1}{x}+\frac{2}{\sqrt{x}}-1}{1+\frac{3}{x}}=-1\)

\(b=\lim\limits_{x\rightarrow+\infty}\frac{1+\frac{3}{x^2}-\frac{1}{x^3}}{\frac{1}{\sqrt{x}}+\frac{1}{x^2}}=\frac{1}{0}=+\infty\)

\(c=\lim\limits_{x\rightarrow-\infty}\frac{1-2\sqrt{\frac{1}{x^2}-\frac{1}{x}}}{\frac{1}{x}-1}=\frac{1}{-1}=-1\)

Bài 2:

\(a=\lim\limits_{x\rightarrow0}\frac{1-cosx}{1-cos3x}=\lim\limits_{x\rightarrow0}\frac{sinx}{3sin3x}=\lim\limits_{x\rightarrow0}\frac{\frac{sinx}{x}}{9.\frac{sin3x}{3x}}=\frac{1}{9}\)

\(b=\lim\limits_{x\rightarrow0}\frac{cotx-sinx}{x^3}=\frac{\infty}{0}=+\infty\)

\(c=\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}\)

Mà \(\left|sinx\right|\le1\Rightarrow\left|\frac{sinx}{2x}\right|\le\frac{1}{\left|2x\right|}\)

Mà \(\lim\limits_{x\rightarrow\infty}\frac{1}{2\left|x\right|}=0\Rightarrow\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}=0\)

Tôi chẳng thể hiểu nổi

\(a=\lim\limits_{x\rightarrow-3}\frac{x^2+2x-3}{x\left(x+3\right)\left(x-\sqrt{3-2x}\right)}=\lim\limits_{x\rightarrow-3}\frac{\left(x-1\right)\left(x+3\right)}{x\left(x+3\right)\left(x-\sqrt{3-2x}\right)}=\lim\limits_{x\rightarrow-3}\frac{x-1}{x\left(x-\sqrt{3-2x}\right)}=-\frac{2}{9}\)

\(b=\lim\limits_{x\rightarrow0}\frac{\sqrt{x+9}-3+\sqrt{x+16}-4}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{x}{\sqrt{x+9}+3}+\frac{x}{\sqrt{x+16}+4}}{x}=\lim\limits_{x\rightarrow0}\left(\frac{1}{\sqrt{x+9}+3}+\frac{1}{\sqrt{x+16}+4}\right)=\frac{7}{24}\)

\(c=\lim\limits_{x\rightarrow\frac{1}{2}}\frac{8x^2-1}{6x^2-5x+1}\) ko phải dạng vô định, đề bài là \(8x^2\) hay \(8x^3\) bạn?

\(d=\lim\limits_{x\rightarrow0}\frac{\left(\sqrt{x^2+1}-1\right)\left(\sqrt{x^2+1}+1\right)\left(4+\sqrt{x^2+16}\right)}{\left(4-\sqrt{x^2+16}\right)\left(4+\sqrt{x^2+16}\right)\left(\sqrt{x^2+1}+1\right)}\)

\(=\lim\limits_{x\rightarrow0}\frac{x^2\left(4+\sqrt{x^2+16}\right)}{-x^2\left(\sqrt{x^2+1}+1\right)}=\lim\limits_{x\rightarrow0}\frac{4+\sqrt{x^2+16}}{-\sqrt{x^2+1}-1}=\frac{8}{-2}=-4\)

Đặt \(t=x-\dfrac{\pi}{4}\), khi đó:

\(\lim\limits_{x\rightarrow\dfrac{\pi}{4}}\dfrac{\sqrt{2}cosx-1}{\sqrt{2}sinx-1}=\lim\limits_{t\rightarrow0}\dfrac{\sqrt{2}cos\left(t+\dfrac{\pi}{4}\right)-1}{\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)-1}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{cost-sint-1}{cost+sint-1}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{1-2sin^2\dfrac{t}{2}-2sin\dfrac{t}{2}.cos\dfrac{t}{2}-1}{1-2sin^2\dfrac{t}{2}+2sin\dfrac{t}{2}.cos\dfrac{t}{2}-1}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{-2sin\dfrac{t}{2}\left(sin\dfrac{t}{2}+cos\dfrac{t}{2}\right)}{-2sin\dfrac{t}{2}\left(sin\dfrac{t}{2}-cos\dfrac{t}{2}\right)}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{sin\dfrac{t}{2}+cos\dfrac{t}{2}}{sin\dfrac{t}{2}-cos\dfrac{t}{2}}\)

\(=-1\)

L'Hospital đi em

lim đề bài cho = \(\lim\limits_{x->\dfrac{\pi}{4}}\dfrac{-\sqrt{2}sinx}{\sqrt{2}cosx}\)

Thay x vào là xong

a) Áp dụng giới hạn một bên thường dùng, ta có : \(\mathop {\lim }\limits_{x \to {4^ + }} \frac{1}{{x - 4}} =  + \infty \)

b) \(\mathop {\lim }\limits_{x \to {2^ + }} \frac{x}{{2 - x}} = \mathop {\lim }\limits_{x \to {2^+ }} \frac{{ - x}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \left( { - x} \right).\mathop {\lim }\limits_{x \to {2^ + }} \frac{1}{{x - 2}}\)

Ta có: \(\mathop {\lim }\limits_{x \to {2^ + }} \left( { - x} \right) =  - \mathop {\lim }\limits_{x \to {2^ + }} x =  - 2;\mathop {\lim }\limits_{x \to {2^ +}} \frac{1}{{x - 2}} =  +\infty \)

\( \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} \frac{x}{{2 - x}} =  - \infty \)

a) Ta có \(t = \frac{1}{x},\) nên khi x tiến đến 0 thì t tiến đến dương vô cùng do đó

\(\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\frac{1}{x}}} = \mathop {\lim }\limits_{t \to  + \infty } {\left( {1 + \frac{1}{t}} \right)^t} = e\)

b) \(\ln y = \ln {\left( {1 + x} \right)^{\frac{1}{x}}} = \frac{1}{x}\ln \left( {1 + x} \right)\)

\(\mathop {\lim }\limits_{x \to 0} \ln y = \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{x} = 1\)

c) \(t = {e^x} - 1 \Leftrightarrow {e^x} = t + 1 \Leftrightarrow x = \ln \left( {t + 1} \right)\)

\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = \mathop {\lim }\limits_{t \to 0} \frac{t}{{\ln \left( {t + 1} \right)}} = 1\)

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) \(\mathop {\lim }\limits_{x \to 2} \frac{{3 - \sqrt {x + 7} }}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {3 - \sqrt {x + 7} } \right)\left( {3 + \sqrt {x + 7} } \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{{3^2} - \left( {x + 7} \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}}\)

                                         \( = \mathop {\lim }\limits_{x \to 2} \frac{{2 - x}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{ - \left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{ - 1}}{{3 + \sqrt {x + 7} }}\)

                                         \( = \frac{{\mathop {\lim }\limits_{x \to 2} \left( { - 1} \right)}}{{\mathop {\lim }\limits_{x \to 2} 3 + \sqrt {\mathop {\lim }\limits_{x \to 2} x + \mathop {\lim }\limits_{x \to 2} 7} }} = \frac{{ - 1}}{{3 + \sqrt {2 + 7} }} =  - \frac{1}{6}\)