A Calculus Problem from Aly Ahmed 200716a

$\begin{aligned} L & = \lim_{x \to 0} \frac {x^2 - \sin^2 x}{x^2 - \tan^2 x} \\ & = \lim_{x \to 0} \frac {(x - \sin x)(x+\sin x)}{(x- \tan x)(x+\tan x)} & \small \blue{\text{By Maclaurin series}} \\ & = \lim_{x \to 0} \frac {\left(x - \left(x - \frac {x^3}6+\frac {x^5}{120}-\cdots\right)\right) \left(x + \left(x - \frac {x^3}6+\frac {x^5}{120}-\cdots\right)\right)}{\left(x - \left(x +\frac {x^3}3+\frac {2x^5}{15}+\cdots\right)\right) \left(x + \left(x + \frac {x^3}3+\frac {2x^5}{15} +\cdots\right)\right)} \\ & = \lim_{x \to 0} \frac {\left(\frac {x^3}6 - \frac {x^5}{120}+ \frac {x^7}{5040} - \cdots\right) \left(2x - \frac {x^3}6+\frac {x^5}{120}-\cdots\right)}{\left(-\frac {x^3}3 - \frac {2x^5}{15} - \frac {17x^7}{315} - \cdots\right) \left(2x + \frac {x^3}3 +\frac {2x^5}{15} + \cdots\right)} & \small \blue{\text{Divide up and down by }x^4} \\ & = \lim_{x \to 0} \frac {\left(\frac 16 - \frac {x^2}{120}+ \frac {x^5}{5040} - \cdots\right) \left(2 - \frac {x^2}6+\frac {x^4}{120}-\cdots\right)}{\left(-\frac 13 - \frac {2x^2}{15} - \frac {17x^4}{315} - \cdots\right) \left(2 + \frac {x^3}3+\frac {2x^4}{15} + \cdots\right)} \\ & = \frac {\frac 16 \times 2}{-\frac 13 \times 2} = - \frac 12 = \boxed{-0.5} \end{aligned}$

In the given limit, the expression assumes a $\dfrac 00$ form. So applying L'Hospital's rule repeatedly we get the limit as

$\displaystyle \lim_{x \to 0} \dfrac {-2}{\sec^2 x+3\sec^4 x}=\boxed {-0.5}$ .