A calculus problem by أحمد الحلاق

$\begin{aligned} L & = \lim_{x \to 0} \frac {5^{3x}-5^x -5^{2x}+1}{\sqrt{1+\cos x}-\sqrt 2} & \small \color{#3D99F6}{\text{This is a 0/0 case, so L'Hôpital's rule applies.}} \\ & = \lim_{x \to 0} \frac {3\ln 5 \cdot 5^{3x}- \ln 5 \cdot 5^x - 2\ln 5 \cdot 5^{2x}}{\frac 12 (1+\cos x)^{-\frac12}(-\sin x)} & \small \color{#3D99F6}{\text{Differentiate up and down w.r.t. }x} \\ & = \lim_{x \to 0} \frac {9\ln^2 5 \cdot 5^{3x}- \ln^2 5 \cdot 5^x - 4\ln^2 5 \cdot 5^{2x}}{-\frac 14 (1+\cos x)^{-\frac32}\sin^2 x - \frac 12 (1+\cos x)^{-\frac12} \cos x} & \small \color{#3D99F6}{\text{Differentiate up and down again.}} \\ & = -8\sqrt 2 \ln^2 5 \approx \boxed{-29.3} \end{aligned}$

Recall that $\displaystyle \lim_{f(x) \to 0} \dfrac{a^{f(x)} - 1}{f(x)} = \ln(a)$ and $\displaystyle \lim_{f(x) \to 0} \dfrac{f(x)}{\sin(f(x))} = 1$ . Now

$\begin{aligned} L &= \displaystyle \lim_{x \to 0} \dfrac{5^{3x} - 5^x - 5^{2x} +1}{\sqrt{1+ \cos x} - \sqrt{2}} \\ &= \displaystyle \lim_{x \to 0} \dfrac{(5^x -1)(5^{2x} - 1)}{\sqrt{1+ \cos x} - \sqrt{2}} \\ &= \displaystyle \lim_{x \to 0} \dfrac{5^x - 1}{x} \cdot \dfrac{5^{2x} -1}{2x} \cdot \dfrac{2x^2}{\sqrt{1+ \cos x} - \sqrt{2}} \\ &= \ln 5 \cdot \ln 5 \cdot \displaystyle \lim_{x \to 0} \dfrac{2x^2}{\sqrt{1+ \cos x} - \sqrt{2}} & \small\color{#3D99F6}{\text{Apply L'Hopital's Rule}} \\ &= \ln^2 5 \cdot \displaystyle \lim_{x \to 0} 4 \cdot \dfrac{x}{-\sin x} \cdot 2 \sqrt{1+\cos x} \\ &= - 8 \sqrt{2} \ln^2 5 = \boxed{-29.3} \end{aligned}$