\(\lim\limits_{x\rightarrow a}\frac{sin\left(\frac{x-a}{2}\right)}{\frac{x-a}{2}}.cos\left(\frac{x+a}{2}\right)=1.cos\left(\frac{a+a}{2}\right)=cosa\)

b/ \(\lim\limits_{x\rightarrow\pi}\frac{sin\frac{\pi}{2}-sin\frac{x}{2}}{\pi-x}=\lim\limits_{x\rightarrow\pi}\frac{sin\left(\frac{\pi-x}{4}\right)}{\frac{\pi-x}{4}}.\frac{cos\left(\frac{\pi+x}{4}\right)}{2}=\frac{cos\left(\frac{\pi+\pi}{4}\right)}{2}=0\)

c/ Đặt \(x-\frac{\pi}{3}=a\Rightarrow x=a+\frac{\pi}{3}\)

\(\lim\limits_{a\rightarrow0}\frac{sina}{1-2cos\left(a+\frac{\pi}{3}\right)}=\lim\limits_{a\rightarrow0}\frac{sina}{1-cosa+\sqrt{3}sina}\)

\(=\lim\limits_{a\rightarrow0}\frac{2sin\frac{a}{2}cos\frac{a}{2}}{-2sin^2\frac{a}{2}+2\sqrt{3}sin\frac{a}{2}cos\frac{a}{2}}=\lim\limits_{a\rightarrow0}\frac{cos\frac{a}{2}}{-sin\frac{a}{2}+\sqrt{3}cos\frac{a}{2}}=\frac{1}{\sqrt{3}}\)

d/Ta có: \(tana-tanb=\frac{sina}{cosa}-\frac{sinb}{cosb}=\frac{sina.cosb-cosa.sinb}{cosa.cosb}=\frac{sin\left(a-b\right)}{cosa.cosb}\)
Áp dụng:

\(\lim\limits_{x\rightarrow a}\frac{\left(tanx-tana\right)\left(tanx+tana\right)}{\frac{sin\left(x-a\right)}{cos\left(x-a\right)}}=\lim\limits_{x\rightarrow a}\frac{sin\left(x-a\right)\left(tanx+tana\right).cos\left(x-a\right)}{sin\left(x-a\right).cosx.cosa}=\lim\limits_{x\rightarrow a}\frac{\left(tanx+tana\right).cos\left(x-a\right)}{cosx.cosa}\)

\(=\frac{2tana}{cos^2a}\)

\(\lim_{x\to-\infty}\left(-x^3+x^2-x+1\right)\)

\(=\lim_{x\to-\infty}\left\lbrack x^3\left(-1+\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}\right)\right\rbrack\)

=+∞ vì \(\lim_{x\to-\infty}x^3=-\) ∞ và \(\lim_{x\to-\infty}-1+\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}=-1<0\)

Tôi chẳng thể hiểu nổi

Nó ko phải dạng vô định thì bạn cứ thay vô và tính toán bình thường

\(\lim\limits_{x\rightarrow0}\dfrac{1-\dfrac{1}{x}}{1+\dfrac{1}{x}}=\lim\limits_{x\rightarrow0}\dfrac{x-1}{x+1}=\dfrac{0-1}{0+1}=-1\)

em thắc mắc chỗ 1/x ạ, nếu x=0 thì không chia được ạ :(((

Đặt t=1-x

=>x=1-t

Khi x->1 thì t->0

\(\frac{\pi x}{2} = \frac{\pi(1 - t)}{2} = \frac{\pi}{2} - \frac{\pi t}{2}\)

Ta có: \(\tan\left(\frac{\pi}{2} - \frac{\pi t}{2}\right) = \cot\left(\frac{\pi t}{2}\right) = \frac{1}{\tan\left(\frac{\pi t}{2}\right)}\)

=>\(L = \lim_{t \to 0} \left[ t \cdot \cot\left(\frac{\pi t}{2}\right) \right] = \lim_{t \to 0} \frac{t}{\tan\left(\frac{\pi t}{2}\right)}\)

\(=\lim_{t\to0}\left(\frac{2}{\pi}\cdot\frac{\frac{\pi t}{2}}{\tan\left(\frac{\pi t}{2}\right)}\right)\)

\(\lim_{u\to0}\frac{\tan u}{u}=1\)

=>\(\lim_{u\to0}\frac{u}{\tan u}=1\) (Với \(u=\frac{\pi t}{2}\) )

\(L = \frac{2}{\pi} \cdot 1 = \frac{2}{\pi}\)

1. Ta có : \(\lim\limits_{x\rightarrow0}\frac{\tan ax}{\tan bx}=\lim\limits_{x\rightarrow0}\left(\frac{\sin ax}{\sin bx}.\frac{\cos ax}{\cos bx}\right)=\lim\limits_{x\rightarrow0}\frac{\sin ax}{\sin bx}=\lim\limits_{x\rightarrow0}\left(\frac{\frac{\sin ax}{ax}}{\frac{\sin bx}{bx}}.\frac{ax}{bx}\right)=\frac{a}{b}\frac{\lim\limits_{x\rightarrow0}\frac{\sin ax}{ax}}{\lim\limits_{x\rightarrow0}\frac{\sin bx}{bx}}=\frac{a}{b}\frac{\lim\limits_{y\rightarrow0}\frac{\sin y}{y}}{\lim\limits_{z\rightarrow0}\frac{\sin z}{z}}=\frac{a}{b}\)

2. Ta có : \(\lim\limits_{x\rightarrow0}\frac{1-\cos ax}{x^2}=\lim\limits_{x\rightarrow0}\frac{2\sin^2\frac{ax}{2}}{x^2}=\lim\limits_{x\rightarrow0}\left[\left(\frac{\sin\frac{ax}{2}.\sin\frac{ax}{2}}{\frac{ax}{2}.\frac{ax}{2}}\right).\frac{a^2}{2}\right]\)

                                   \(=\frac{a^2}{2}\left(\lim\limits_{y\rightarrow0}\frac{\sin y}{y}\right)^2=\frac{a^2}{2}\)

a/ \(\lim\limits_{x\rightarrow-1}\dfrac{2x^3-5x-4}{\left(x+1\right)^2}=\dfrac{2.\left(-1\right)^3-5\left(-1\right)-4}{\left(-1+1\right)^2}=-\dfrac{1}{0}=-\infty\)

b/ \(\lim\limits\left(x^3+2\sqrt{x^5}-1\right)=\lim\limits x^3\left(1+0-0\right)=+\infty\)

giúp em câu này với ạ https://hoc24.vn/hoi-dap/tim-kiem?id=353722985710&q=lim%C2%A0\(\dfrac{1-\dfrac{1}{x}}{1+\dfrac{1}{x}}\)%C2%A0khi+x+ti%E1%BA%BFn+t%E1%BB%9Bi+0

\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+x}-\sqrt[3]{x^3-x^2}\right)\)

\(=\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+x}-x+x-\sqrt[3]{x^3-x^2}\right)\)

\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{x}{\sqrt{x^2+x}+x}+\dfrac{x^2}{x^2+x.\sqrt[3]{x^3-x^2}+\sqrt[3]{\left(x^3-x^2\right)^2}}\right)\)

\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{1}{\sqrt{1+\dfrac{1}{x}}+1}+\dfrac{1}{1+\sqrt[3]{1-\dfrac{1}{x}}+\sqrt[3]{\left(1-\dfrac{1}{x}\right)^2}}\right)\)

\(=\dfrac{1}{\sqrt{1+0}+1}+\dfrac{1}{1+\sqrt[3]{1-0}+\sqrt[3]{\left(1-0\right)^2}}\)

\(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)