Limited Trigonometry - III

Calculus Level 5

f ( x ) = tan ( x 2 ) sec ( x ) + tan ( x 2 2 ) sec ( x 2 ) + + tan ( x 2 n ) sec ( x 2 n 1 ) f(x)=\tan\left( \frac{x}{2} \right) \cdot \sec \left( x \right)+\tan\left( \frac{x}{2^2} \right) \cdot \sec \left( \frac{x}{2} \right)+\ldots+\tan\left( \frac{x}{2^n} \right) \cdot \sec \left( \frac{x}{2^{n-1}} \right) g ( x ) = f ( x ) + tan ( x 2 n ) g(x)=f(x)+\tan\left( \frac{x}{2^n} \right)

where x ( π 2 , π 2 ) x \in \left( -\frac{\pi}{2} , \frac{\pi}{2} \right) and n N n \in \mathbb{N}

Evaluate the limit : lim x 0 ( g ( x ) x ) 1 x 3 \displaystyle \lim_{x \to 0} \left( \dfrac{g(x)}{x} \right)^{\frac{1}{x^3}}


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e e e 1 3 e^{\frac{1}{3}} Limit does not exist 1 None of the above

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1 solution

Ashwin Gopal
May 21, 2015

G(x)=tan(x). For proof check the previous problem (bcs I hate latex,sorry)so apply the limit u will that limit as e^infinity so not possible

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