26 câu Dạng 6: Tìm giới hạn hàm lượng giác có đáp án

Ta có $A=\underset{x\to 0}{\lim }\frac{2{\sin }^2\frac{ax}{2}}{x^2}=\frac{a^2}{2}\underset{x\to 0}{\lim }{\left(\frac{\sin \frac{ax}{2}}{\frac{ax}{2}}\right)}^2=\frac{a^2}{2}$

Hướng dẫn giải

Ta có $\frac{1-\cos x.\cos 2x.\cos 3x}{x^2}$

$=\frac{1-\cos x+\cos x\cos 2x\left(1-\cos 3x\right)+\cos x\left(1-\cos 2x\right)}{x^2}$

$=\frac{1-\cos x}{x^2}+\cos x.\cos 2x\frac{1-\cos 3x}{x^2}+\cos \frac{1-\cos 2x}{x^2}$

$B=\underset{x\to 0}{\lim }\left(\frac{1-\cos x}{x^2}+\cos x.\cos 2x\frac{1-\cos 3x}{x^2}+\cos x\frac{1-\cos 2x}{x^2}\right)=7$

Hướng dẫn giải

Ta có $\frac{1+\sin x-\cos x}{1+\sin 2x-\cos 2x}=\frac{2{\sin }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\sin }^{2}x+2\sin x\cos x}$

$\Rightarrow A=\underset{x\to 0}{\lim }\frac{1}{2}.\frac{\sin \frac{x}{2}}{\frac{x}{2}}.\frac{x}{\sin x}.\frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\sin x+\cos x}=\frac{1}{2}$

Hướng dẫn giải

.

$C=\underset{x\to 0}{\lim }\frac{{\tan }^{2}2x}{1-\sqrt[3]{\cos 2x}}=\underset{x\to 0}{\lim }\frac{{\tan }^{2}2x\left(1+\sqrt[3]{\cos 2x}+\sqrt[3]{{\cos }^{2}2x}\right)}{1-\cos 2x}$

$=\underset{x\to 0}{\lim }\frac{{\tan }^{2}2x\left(1+\sqrt[3]{\cos 2x}+\sqrt[3]{{\cos }^{2}2x}\right)}{2{\sin }^{2}x}$

$=2\underset{x\to 0}{\lim }{\left(\frac{\tan 2x}{2x}\right)}^2.{\left(\frac{x}{\sin x}\right)}^2.\left(1+\sqrt[3]{\cos 2x}+\sqrt[3]{{\cos }^{2}2x}\right)$

$\Rightarrow C=6$

Hướng dẫn giải

$A=\underset{x\to 1}{\lim }\frac{\sin \pi \left(1-x^m\right)}{\sin \pi \left(1-x^n\right)}=\underset{x\to 1}{\lim }\frac{\sin \pi \left(1-x^m\right)}{\pi \left(1-x^m\right)}.\underset{x\to 1}{\lim }\frac{\pi \left(1-x^n\right)}{\sin \pi \left(1-x^n\right)}.\underset{x\to 1}{\lim }\frac{1-x^n}{1-x^m}$

$=\underset{x\to 1}{\lim }\frac{1-x^n}{1-x^m}=\underset{x\to 1}{\lim }\frac{\left(1-x\right)\left(x^{n-1}+x^{n-2}+\mathrm{...}+1\right)}{\left(1-x\right)\left(x^{m-1}+x^{m-2}+\mathrm{...}+1\right)}=\frac{n}{m}$

Ta có: $B=\underset{x\to 1}{\lim }\frac{\tan \left(x-1\right)}{x-1}=1$

Ta có: $C=\underset{x\to 0}{\lim }\frac{2\tan 2x}{2x}.\frac{5\sin 5x}{5x}=10$

Ta có $D=\underset{x\to 0}{\lim }\frac{\sin x\left(\cos x-1\right)}{x^3\cos x}=-\underset{x\to 0}{\lim }\frac{2\sin x.{\sin }^2\frac{x}{2}}{x^3\cos x}=-\frac{1}{2}\underset{x\to 0}{\lim }\frac{\sin x}{x}{\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)}^2=-\frac{1}{2}$

Ta có $A=\underset{x\to 0}{\lim }\frac{\cos 3x-\cos 4x}{\cos 5x-\cos 6x}=\underset{x\to 0}{\lim }\frac{\sin \frac{7x}{2}.\sin \frac{x}{2}}{\sin \frac{11x}{2}.\sin \frac{x}{2}}=\underset{x\to 0}{\lim }\frac{\sin \frac{7x}{2}}{\sin \frac{11x}{2}}=\frac{7}{11}$

Ta có $B=\underset{x\to 0}{\lim }\frac{1-\sqrt[3]{1+2\sin 2x}}{\sin 3x}=\underset{x\to 0}{\lim }\frac{-2\sin 2x}{\sin 3x\left(1+\sqrt[3]{1+2\sin 2x}+\sqrt[3]{{\left(1+2\sin 2x\right)}^2}\right)}=-\frac{4}{9}$

Ta có $C=\underset{x\to 0}{\lim }\frac{{\sin }^{2}2x}{\sqrt[3]{\cos x}-\sqrt[4]{\cos x}}=\underset{x\to 0}{\lim }\frac{\frac{{\sin }^{2}2x}{x^2}}{\frac{\sqrt[3]{\cos x}-1}{x^2}+\frac{1-\sqrt[4]{\cos x}}{x^2}}$

$\underset{x\to 0}{\lim }\frac{1-\sqrt[3]{\cos x}}{x^2}=\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2\left(1+\sqrt[3]{\cos 2x}+\sqrt[3]{{\cos }^{2}2x}\right)}=\underset{x\to 0}{\lim }\frac{2{\sin }^2\frac{x}{2}}{x^2\left(1+\sqrt[3]{\cos 2x}+\sqrt[3]{{\cos }^{2}2x}\right)}=\frac{1}{6}$

$\underset{x\to 0}{\lim }\frac{1-\sqrt[4]{\cos x}}{x^2}=\underset{x\to 0}{\lim }\frac{\left(1-\cos x\right)}{x^2\left(1+\sqrt[4]{\cos x}\right)\left(1+\sqrt{\cos x}\right)}=\frac{1}{8}$

$\underset{x\to 0}{\lim }\frac{{\sin }^{2}2x}{x^2}=4$

Vậy $C=\frac{4}{-\frac{1}{6}+\frac{1}{8}}=-96$

Ta có $D=\underset{x\to 0}{\lim }\frac{{\sin }^{4}2x}{{\sin }^{4}3x}=\underset{x\to 0}{\lim }\frac{\frac{{\sin }^{4}2x}{x^4}}{\frac{{\sin }^{4}3x}{x^2}}=\frac{16}{81}$

Ta có $E=\underset{x\to 0}{\lim }\frac{\frac{1-\sin \left(\frac{\pi }{2}\cos x\right)}{\tan x}}{\frac{\sin \left(\tan x\right)}{\tan x}}$ mà $\underset{x\to 0}{\lim }\frac{\sin \left(\tan x\right)}{\tan x}=1$

Lại có $\underset{x\to 0}{\lim }\frac{1-\sin \left(\frac{\pi }{2}\cos x\right)}{\tan x}=\underset{x\to 0}{\lim }\frac{1-\cos \left[\frac{\pi }{2}\left(1-\cos x\right)\right]}{\tan x}=\underset{x\to 0}{\lim }\frac{2{\sin }^2\left(\frac{\pi {\sin }^2\frac{x}{2}}{2}\right)}{\tan x}$

$=\frac{\pi }{4}\underset{x\to 0}{\lim }\frac{{\sin }^2\left(\frac{\pi {\sin }^2\frac{x}{2}}{2}\right)}{\frac{\pi {\sin }^2\frac{x}{2}}{2}}.\frac{{\sin }^2\frac{x}{2}}{{\left(\frac{x}{2}\right)}^2}.x.\frac{x}{\tan x}=0$

Do đó E= 0

Ta có $0\le \left|\cos \frac{2}{nx}\right|\le 1\iff 0\le \left|x^2\cos \frac{2}{nx}\right|\le x^2$ mà $\underset{x\to 0}{\lim }{x}^2=0$ nên $\underset{x\to 0}{\lim }{x}^2\cos \frac{2}{nx}=0$

Ta có $L=\underset{x\to \frac{\pi }{2}}{\lim }\frac{\cos x}{x-\frac{\pi }{2}}=\underset{x\to \frac{\pi }{2}}{\lim }\frac{\sin \left(\frac{\pi }{2}-x\right)}{x-\frac{\pi }{2}}=-1$

$H=\underset{x\to 0}{\lim }\frac{\sqrt[m]{\cos ax}-\sqrt[m]{\cos bx}}{{\sin }^{2}x}=\underset{x\to 0}{\lim }\frac{\sqrt[m]{\cos ax}-1+1-\sqrt[m]{\cos bx}}{{\sin }^{2}x}$

$=\underset{x\to 0}{\lim }\frac{\cos ax-1}{\left[{\left(\sqrt[m]{\cos ax}\right)}^{m-1}+{\left(\sqrt[m]{\cos ax}\right)}^{m-2}+\mathrm{...}+1\right]{\sin }^{2}x}-\underset{x\to 0}{\lim }\frac{\cos bx-1}{\left[{\left(\sqrt[m]{\cos bx}\right)}^{m-1}+{\left(\sqrt[m]{\cos bx}\right)}^{m-2}+\mathrm{...}+1\right]{\sin }^{2}x}$

$=\underset{x\to 0}{\lim }\frac{2{\sin }^2\frac{bx}{2}}{\left[{\left(\sqrt[m]{\cos bx}\right)}^{m-1}+{\left(\sqrt[m]{\cos bx}\right)}^{m-2}+\mathrm{...}+1\right]{\sin }^{2}x}-\underset{x\to 0}{\lim }\frac{2{\sin }^2\frac{ax}{2}}{\left[{\left(\sqrt[m]{\cos ax}\right)}^{m-1}+{\left(\sqrt[m]{\cos ax}\right)}^{m-2}+\mathrm{...}+1\right]{\sin }^{2}x}$

$=\underset{x\to 0}{\lim }\frac{\frac{b^2}{2}.\frac{{\sin }^2\frac{bx}{2}}{\frac{b^2{x}^2}{4}}}{\left[{\left(\sqrt[m]{\cos bx}\right)}^{m-1}+{\left(\sqrt[m]{\cos bx}\right)}^{m-2}+\mathrm{...}+1\right]\frac{{\sin }^{2}x}{x^2}}-\underset{x\to 0}{\lim }\frac{\frac{a^2}{2}.\frac{{\sin }^2\frac{ax}{2}}{\frac{a^2{x}^2}{4}}}{\left[{\left(\sqrt[m]{\cos ax}\right)}^{m-1}+{\left(\sqrt[m]{\cos ax}\right)}^{m-2}+\mathrm{...}+1\right]\frac{{\sin }^{2}x}{x^2}}$

$=\frac{b^2-a^2}{2m}$

Ta có $M=\underset{x\to 0}{\lim }\frac{1-\sqrt[n]{\cos ax}}{x^2}=\underset{x\to 0}{\lim }\frac{1-\cos ax}{\left[{\left(\sqrt[n]{\cos ax}\right)}^{n-1}+{\left(\sqrt[n]{\cos ax}\right)}^{n-2}+\mathrm{...}+1\right]x^2}$

$=\underset{x\to 0}{\lim }\frac{\frac{a^2}{2}{\sin }^2\frac{ax}{2}}{\left[{\left(\sqrt[n]{\cos ax}\right)}^{n-1}+{\left(\sqrt[n]{\cos ax}\right)}^{n-2}+\mathrm{...}+1\right]\frac{a^2{x}^2}{4}}=\frac{a^2}{2n}$

Ta có $M=\underset{x\to 0}{\lim }\frac{\frac{\sqrt[3]{3x+1}-\sqrt{2x+1}}{x^2}}{\frac{1-\cos 2x}{x^2}}=\underset{x\to 0}{\lim }\frac{\frac{\sqrt[3]{3x+1}-\left(x+1\right)+\left(x+1\right)-\sqrt{2x+1}}{x^2}}{\frac{2{\sin }^{2}x}{x^2}}$

$=\underset{x\to 0}{\lim }\frac{\frac{\sqrt[3]{3x+1}-x-1}{x^2}}{\frac{2{\sin }^{2}x}{x^2}}+\underset{x\to 0}{\lim }\frac{\frac{x+1-\sqrt{2x+1}}{x^2}}{\frac{2{\sin }^{2}x}{x^2}}$

$=\underset{x\to 0}{\lim }\frac{\frac{-x^3-3x^2}{x^2\left({\left(\sqrt[3]{3x+1}\right)}^2+\left(x+1\right)\sqrt[3]{3x+1}+{\left(x+1\right)}^2\right)}}{\frac{2{\sin }^{2}x}{x^2}}+\underset{x\to 0}{\lim }\frac{\frac{x^2}{x^2\left(x+1+\sqrt{2x+1}\right)}}{\frac{2{\sin }^{2}x}{x^2}}$

$=\frac{-1}{2}+\frac{1}{4}=-\frac{1}{4}\Rightarrow a=1;b=4\Rightarrow a+b=5$

Ta có $\underset{x\to 0}{\lim }\frac{2\sqrt{1+x}-\sqrt[3]{8-x}}{\sin 3x}=\underset{x\to 0}{\lim }\frac{2\sqrt{1+x}-2+2-\sqrt[3]{8-x}}{\sin 3x}=\underset{x\to 0}{\lim }\frac{2\sqrt{1+x}-2}{\sin 3x}+\underset{x\to 0}{\lim }\frac{2-\sqrt[3]{8-x}}{\sin 3x}$

$=\underset{x\to 0}{\lim }\frac{\frac{2x}{1+\sqrt{1+x}}}{3x\frac{\sin 3x}{3x}}+\underset{x\to 0}{\lim }\frac{\frac{x}{4+2\sqrt[3]{8-x}+{\left(\sqrt[3]{8-x}\right)}^2}}{3x\frac{\sin 3x}{3x}}=\underset{x\to 0}{\lim }\frac{\frac{2}{1+\sqrt{1+x}}}{3\frac{\sin 3x}{3x}}+\underset{x\to 0}{\lim }\frac{\frac{1}{4+2\sqrt[3]{8-x}+{\left(\sqrt[3]{8-x}\right)}^2}}{3\frac{\sin 3x}{3x}}$

$=\frac{1}{3}+\frac{1}{36}=\frac{13}{36}=\frac{a}{b}\Rightarrow a+b=49$