Bài tập nâng cao · Bài 26
Bài tập nâng cao Giới hạn của hàm số · Bài 26
Tìm giới hạn
a. $\lim\limits_{x\to 0}\frac{1-\cos 4x}{2x^{2}}$
b. $\lim\limits_{x\to 0}\frac{\sin^{2}x}{2x^{2}}$
c. $\lim\limits_{x\to 0}\frac{1-\cos 3x}{1-\cos 5x}$
d. $\lim\limits_{x\to 0}\frac{3-\cos x-\cos 2x-\cos 3x}{1-\cos x}$
a. $\lim\limits_{x\to 0}\frac{1-\cos 4x}{2x^{2}}$
b. $\lim\limits_{x\to 0}\frac{\sin^{2}x}{2x^{2}}$
c. $\lim\limits_{x\to 0}\frac{1-\cos 3x}{1-\cos 5x}$
d. $\lim\limits_{x\to 0}\frac{3-\cos x-\cos 2x-\cos 3x}{1-\cos x}$
Xem lời giải
Lời giải
a. $\lim\limits_{x\to 0}\frac{1-\cos 4x}{2x^{2}}=\lim\limits_{x\to 0}\frac{2\sin^{2}2x}{2x^{2}}=\lim\limits_{x\to 0}4.\left( \frac{\sin 2x}{2x} \right)^{2}=4$.
b. $\lim\limits_{x\to 0}\frac{\sin^{2}x}{2x^{2}}=\lim\limits_{x\to 0}\frac{1}{2}\left( \frac{\sin x}{x} \right)^{2}=\frac{1}{2}$.
c. $\lim\limits_{x\to 0}\frac{1-\cos 3x}{1-\cos 5x}=\lim\limits_{x\to 0}\frac{\frac{1-\cos 2\frac{3x}{2}}{2}}{\frac{1-\cos 2\frac{5x}{2}}{2}}=\lim\limits_{x\to 0}\frac{\sin^{2}\frac{3x}{2}}{\sin^{2}\frac{5x}{2}}=\lim\limits_{x\to 0}\left( \frac{\sin^{2}\frac{3x}{2}\left( \frac{5x}{2} \right)^{2}}{\left( \frac{3x}{2} \right)^{2}\sin^{2}\frac{5x}{2}}.\frac{9}{25} \right)=\frac{9}{25}$.
d. $\lim\limits_{x\to 0}\frac{3-\cos x-\cos 2x-\cos 3x}{1-\cos x}=\lim\limits_{x\to 0}\left( 1+\frac{1-\cos 2x}{1-\cos x}+\frac{1-\cos 3x}{1-\cos x} \right)$
$=1+\lim\limits_{x\to 0}\frac{\sin^{2}x}{\sin^{2}\frac{x}{2}}+\lim\limits_{x\to 0}\frac{\sin^{2}\frac{3x}{2}}{\sin^{2}\frac{x}{2}}=1+\lim\limits_{x\to 0}\left( \frac{\sin^{2}x}{x^{2}}.\frac{\left( \frac{x}{2} \right)^{2}}{\sin^{2}\frac{x}{2}}.4 \right)+\lim\limits_{x\to 0}\left( \frac{\sin^{2}\frac{3x}{2}}{\left( \frac{3x}{2} \right)^{2}}.\frac{\left( \frac{x}{2} \right)^{2}}{\sin^{2}\frac{x}{2}}.9 \right)$
$=1+4+9=14$
b. $\lim\limits_{x\to 0}\frac{\sin^{2}x}{2x^{2}}=\lim\limits_{x\to 0}\frac{1}{2}\left( \frac{\sin x}{x} \right)^{2}=\frac{1}{2}$.
c. $\lim\limits_{x\to 0}\frac{1-\cos 3x}{1-\cos 5x}=\lim\limits_{x\to 0}\frac{\frac{1-\cos 2\frac{3x}{2}}{2}}{\frac{1-\cos 2\frac{5x}{2}}{2}}=\lim\limits_{x\to 0}\frac{\sin^{2}\frac{3x}{2}}{\sin^{2}\frac{5x}{2}}=\lim\limits_{x\to 0}\left( \frac{\sin^{2}\frac{3x}{2}\left( \frac{5x}{2} \right)^{2}}{\left( \frac{3x}{2} \right)^{2}\sin^{2}\frac{5x}{2}}.\frac{9}{25} \right)=\frac{9}{25}$.
d. $\lim\limits_{x\to 0}\frac{3-\cos x-\cos 2x-\cos 3x}{1-\cos x}=\lim\limits_{x\to 0}\left( 1+\frac{1-\cos 2x}{1-\cos x}+\frac{1-\cos 3x}{1-\cos x} \right)$
$=1+\lim\limits_{x\to 0}\frac{\sin^{2}x}{\sin^{2}\frac{x}{2}}+\lim\limits_{x\to 0}\frac{\sin^{2}\frac{3x}{2}}{\sin^{2}\frac{x}{2}}=1+\lim\limits_{x\to 0}\left( \frac{\sin^{2}x}{x^{2}}.\frac{\left( \frac{x}{2} \right)^{2}}{\sin^{2}\frac{x}{2}}.4 \right)+\lim\limits_{x\to 0}\left( \frac{\sin^{2}\frac{3x}{2}}{\left( \frac{3x}{2} \right)^{2}}.\frac{\left( \frac{x}{2} \right)^{2}}{\sin^{2}\frac{x}{2}}.9 \right)$
$=1+4+9=14$