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toi_yeu_vn
07-11-2008, 09:49 PM
Tìm giới hạn sau:

\lim_{x \to 0} \left ( \frac{1}{tanx}-\frac{1}{sin^2\frac{x}{2}} \right )

tvdaikg
12-11-2008, 12:41 AM
Ta có
\lim_{x \to 0} \left ( \frac{1}{tanx}-\frac{1}{sin^2\frac{x}{2}} \right )= \lim_{x \to 0} \left [ \frac{1}{sin^2\frac{x}{2}}.(\frac{cos x.tan\frac{x}{2}}{2}-1)\right ]
Mà \lim_{x \to 0} \frac{1}{sin^2\frac{x}{2}}=+\infty và
\lim_{x \to 0}(\frac{cos x.tan\frac{x}{2}}{2}-1)=-1.
Do đó, \lim_{x \to 0} \left ( \frac{1}{tanx}-\frac{1}{sin^2\frac{x}{2}} \right ) = -\infty.