Calculating $\lim\limits_{x \to 0} \frac{10^x - 2^x - 5^x + 1}{x \cdot \tan x}$ without L'Hospital

Question

I was trying to solve this problem a while back and I did but as L'Hospital's Rule is not permitted for usage in School Tests, there was no such answer that was able to help me solve this limit and there still isn't on Stack Exchange (Math).

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

Answer 1

We can first simplify it like that:

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

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

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

The highlighted limit is $1$.

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

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

and now we can apply the following formula, which is allowed to be used in NCERT Book:

$$\lim_{x\to0}\frac{a^x-1}{x} = ln(a)$$

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

ln here represents the logarithm of base e.

You can find the proof of the formula here.

Answer 2

$$\begin{aligned}\lim_{x\to0}&\frac{(2^x-1)(5^x-1)}{x\tan(x)}\\=\lim_{x\to0}&\frac{x(2^x-1)(5^x-1)}{x\cdot x\tan(x)}\\=\lim_{x\to0}&\frac{(2^x-1)(5^x-1)}{x^2}\cdot\lim_{x\to0}\frac{x}{\tan(x)}\end{aligned}$$