Question

$\underset{t\to 0}{\mathrm{Lim}}\underset{0}{\overset{2\pi }{\int }}\frac{|\sin (x+t)-\sin x|}{|t|}dx$ equals

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

Answer: (D)

$I=\underset{t\to 0}{\mathrm{Lim}}\underset{0}{\overset{2\pi }{\int }}\left|\frac{\sin (x+t)-\sin x}{t}\right|dx=\underset{0}{\overset{2\pi }{\int }}\left(\underset{t\to 0}{\mathrm{Lim}}\left|\frac{2\cos \left(x+\frac{t}{2}\right)\sin \frac{t}{2}}{t}\right|\right)dx$
$=\underset{0}{\overset{2\pi }{\int }}\underset{t\to 0}{\mathrm{Lim}}\left|\cos \left(x+\frac{t}{2}\right)\frac{\sin \frac{t}{2}}{\frac{t}{2}}\right|=\underset{0}{\overset{2\pi }{\int }}|\cos x|dx=4$