Find the limit, $\lim_{x \to 0}{\frac{10^x-2^x-5^x+1}{x^2}}$(Without L'Hospital rule)
By using L'Hospital rule twice we can easily calculate the limit. But, how could we can calculate this without L'Hospital.
$$\lim_{x \to 0}{\frac{10^x-2^x-5^x+1}{x^2}}\\{=\lim_{x \to 0}{\frac{10^x\ln10-2^x\ln2-5^x\ln5}{2x}}\\=\lim_{x \to 0}{\frac{10^x(\ln10)^2-2^x(\ln2)^2-5^x(\ln5)^2}{2}}\\=\frac{1}{2}((\ln10)^2-(\ln2)^2-(\ln5)^2)\\=\frac{1}2((\ln10+\ln2)(\ln10-\ln2)-(\ln5)^2)\\=\frac{1}{2}\ln5(\ln20-\ln5)\\=\ln5\ln2}$$
We can use $\lim_{x\to 0}{\frac{e^x-1}{x}}=1$to calculate the limit after using L'Hospital rule once.
How can I calculate this without help L'hospital rule? Any help and suggestion is highly solicited.
\begin{gathered} \mathop {\lim }\limits_{x \to 0} \frac{{{{10}^x} - {2^x} - {5^x} + 1}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{5^x}{2^x} - {2^x} - {5^x} + 1}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{{5^x}\left( {{2^x} - 1} \right) - \left( {{2^x} - 1} \right)}}{{{x^2}}} \hfill \\ = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {{5^x} - 1} \right)\left( {{2^x} - 1} \right)}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {{5^x} - 1} \right)}}{x}\frac{{\left( {{2^x} - 1} \right)}}{x} = \ln \left( 5 \right)\ln \left( 2 \right) \hfill \\ \because a > 0,\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^{x\left( {\ln \left( a \right)} \right)}} - 1}}{{x\ln \left( a \right)}}\ln \left( a \right) = \ln \left( a \right) \hfill \\ \end{gathered}