A calculus problem by Aly Ahmed

Similar solution with @Isaac YIU Math Studio 's

$\begin{aligned} L & = \lim_{x \to 0} \frac {x^2 - \sin^2 x}{x^2 - \tan^2 x} \\ & = \color{#3D99F6} \lim_{x \to 0} \frac {x - \sin x}{x - \tan x} \times \lim_{x \to 0} \frac {x + \sin x}{x + \tan x} & \small \color{#3D99F6} \text{Both are 0/0 cases, L'Hôpital's rule applies} \\ & = {\color{#3D99F6} \lim_{x \to 0} \frac {1 - \cos x}{1 - \sec^2 x}} \times \lim_{x \to 0} \frac {1 + \cos x}{1 + \sec^2 x} & \small \color{#3D99F6} \text{Again a 0/0 case} \\ & = {\color{#3D99F6} \lim_{x \to 0} \frac {\sin x}{- 2 \sec^2 x \tan x}} \times \frac 22 & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \lim_{x \to 0} - \frac {\cos^3 x}2 = - \frac 12 = \boxed {-0.5} \end{aligned}$

---

Reference: L'Hôpital's rule

$\lim _{ x\rightarrow 0 }{ \frac { { x }^{ 2 }-\sin ^{ 2 }{ x } }{ { x }^{ 2 }-\tan ^{ 2 }{ x } } } \\ =\lim _{ x\rightarrow 0 }{ \frac { { x }-\sin { x } }{ { x }-\tan { x } } } \times \quad \lim _{ x\rightarrow 0 }{ \frac { { x }+\sin { x } }{ { x }+\tan { x } } } \\ =\lim _{ x\rightarrow 0 }{ \frac { 1-\cos { x } }{ 1-\sec ^{ 2 }{ x } } } \times \quad \lim _{ x\rightarrow 0 }{ \frac { { 1 }+\frac { \sin { x } }{ x } }{ { 1 }+\frac { \tan { x } }{ x } } } \\ =\lim _{ x\rightarrow 0 }{ \frac { 1-\cos { x } }{ \left( 1-\sec { x } \right) \left( 1+\sec { x } \right) } } \\ =\lim _{ x\rightarrow 0 }{ \frac { -\cos { x } }{ 1+\sec { x } } } \\ =-\frac { 1 }{ 2 }$