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受欢迎的微积分 >

limit as x approaches 0 of (x-arctan(x))/(x^2tan(x))

解答

x \to 0 lim (\frac{x - arctan ( x )}{x ^{2} tan ( x )})

解答

\frac{1}{3}

+1

十进制

0.33333 \dots

求解步骤

x \to 0 lim (\frac{x - arctan ( x )}{x ^{2} tan ( x )})

使用洛必达法则

= x \to 0 lim (\frac{1 - \frac{1}{x ^{2} + 1}}{2 x tan ( x ) + sec ^{2} ( x ) x ^{2}})

使用洛必达法则

= x \to 0 lim (\frac{\frac{2 x}{( x ^{2} + 1 ) ^{2}}}{2 x ^{2} sec ^{2} ( x ) tan ( x ) + 4 x sec ^{2} ( x ) + 2 tan ( x )})

化简

\frac{\frac{2 x}{( x ^{2} + 1 ) ^{2}}}{2 x ^{2} sec ^{2} ( x ) tan ( x ) + 4 x sec ^{2} ( x ) + 2 tan ( x )} : \frac{x}{( x ^{2} + 1 ) ^{2} ( x ^{2} sec ^{2} ( x ) tan ( x ) + 2 x sec ^{2} ( x ) + tan ( x ) )}

= x \to 0 lim (\frac{x}{( x ^{2} + 1 ) ^{2} ( x ^{2} sec ^{2} ( x ) tan ( x ) + 2 x sec ^{2} ( x ) + tan ( x ) )})

使用洛必达法则

= x \to 0 lim (\frac{1}{- 2 x ^{6} sec ^{2} ( x ) + 3 x ^{6} sec ^{4} ( x ) + 10 x ^{5} sec ^{2} ( x ) tan ( x ) + 7 x ^{4} sec ^{2} ( x ) + 6 x ^{4} sec ^{4} ( x ) + 4 x ^{3} tan ( x ) + 16 x ^{3} sec ^{2} ( x ) tan ( x ) + 12 x ^{2} sec ^{2} ( x ) + 3 x ^{2} sec ^{4} ( x ) + 4 x tan ( x ) + 6 x sec ^{2} ( x ) tan ( x ) + 3 sec ^{2} ( x )})

代入值

x = 0

= \frac{1}{- 2 \cdot 0 ^{6} sec ^{2} ( 0 ) + 3 \cdot 0 ^{6} sec ^{4} ( 0 ) + 10 \cdot 0 ^{5} sec ^{2} ( 0 ) tan ( 0 ) + 7 \cdot 0 ^{4} sec ^{2} ( 0 ) + 6 \cdot 0 ^{4} sec ^{4} ( 0 ) + 4 \cdot 0 ^{3} tan ( 0 ) + 16 \cdot 0 ^{3} sec ^{2} ( 0 ) tan ( 0 ) + 12 \cdot 0 ^{2} sec ^{2} ( 0 ) + 3 \cdot 0 ^{2} sec ^{4} ( 0 ) + 4 \cdot 0 \cdot tan ( 0 ) + 6 \cdot 0 \cdot sec ^{2} ( 0 ) tan ( 0 ) + 3 sec ^{2} ( 0 )}

化简

\frac{1}{- 2 \cdot 0 ^{6} sec ^{2} ( 0 ) + 3 \cdot 0 ^{6} sec ^{4} ( 0 ) + 10 \cdot 0 ^{5} sec ^{2} ( 0 ) tan ( 0 ) + 7 \cdot 0 ^{4} sec ^{2} ( 0 ) + 6 \cdot 0 ^{4} sec ^{4} ( 0 ) + 4 \cdot 0 ^{3} tan ( 0 ) + 16 \cdot 0 ^{3} sec ^{2} ( 0 ) tan ( 0 ) + 12 \cdot 0 ^{2} sec ^{2} ( 0 ) + 3 \cdot 0 ^{2} sec ^{4} ( 0 ) + 4 \cdot 0 \cdot tan ( 0 ) + 6 \cdot 0 \cdot sec ^{2} ( 0 ) tan ( 0 ) + 3 sec ^{2} ( 0 )} : \frac{1}{3}

= \frac{1}{3}

作图

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