Calculating $\lim\limits_{x \to 0} \frac{10^x - 2^x - 5^x + 1}{x \cdot \tan x}$

Question

I am trying to solve this limit without using Taylor expansions to explain it to a secondary school student

$$\lim_{x \to 0} \frac{10^x - 2^x - 5^x + 1}{x \cdot \tan x}.$$

Answer 1

$$\lim_{x\to0} \frac{(2^x-1)(5^x-1)}{x\tan(x)}$$

$$\lim_{x\to0} \frac{x(2^x-1)(5^x-1)}{x\cdot x\tan(x)}$$

$$\lim_{x\to0} \frac{(2^x-1)(5^x-1)}{x^2}\cdot\color{red}{\lim_{x\to0}\frac{x}{\tan(x)}}$$

The highlighted limit is $1$.

$$\lim_{x\to0} \frac{(2^x-1)(5^x-1)}{x^2}$$

$$\lim_{x\to0}\frac{2^x-1}{x}\cdot\lim_{x\to0}\frac{5^x-1}{x}$$

Which may be individually calculated using L'hospital's rule.