Find the limit by L'hospital rule

Question

Finding limit by l'hospital rule

$$ \lim_{x \to 0} \frac{7^{2x}-5^{3x}}{2x - \arctan 3x} $$

Can anyone to explain me how the L'hospital rule works?

Answer 1

$$\frac{7^{2 x}-5^{3 x}}{2 x-\tan ^{-1}(3 x)}\to\frac{2 \times 7^{2 x} \ln 7 -3 \times 5^{3 x} \ln 5}{2-\displaystyle \frac{3}{9 x^2+1}}$$ No more problem. Compute the limit.

Answer 2

$$L=\lim_{x\to 0}\frac{7^{2x}-5^{3x}}{2x-\tan^{-1}3x}$$ D. up and down separtely w.r.t. $x$ $$L=\lim_{x\to 0} \frac{2.7^{2x} \ln 7 -3.5^{3x}\ln 5}{2-\frac{3}{1+9x^2}}$$ $$L=-\ln(7^2/5^3)=\ln(49/125).$$