Question

Вычислить предел, №2
В числителе не $105^x$, а $10*5^x$

Answer 1

$\displaystyle \lim_{x \to 1+}\frac{5^{2x}-10\cdot 5^x+25}{\sqrt{x^2-1}-\sqrt{2x-1}}=\lim_{x \to 1+}\frac{(5^x-5)^2\cdot (\sqrt{x^2-1}+\sqrt{2x-2})}{(\sqrt{x^2-1}-\sqrt{2x-2})(\sqrt{x^2-1}+\sqrt{2x-2})}$

$=\displaystyle \lim_{x \to 1+}\frac{25(5^{x-1}-1)^2\cdot (\sqrt{x^2-1}+\sqrt{2x-2})}{x^2-1-2x+2}=\\ \\ \\ =\lim_{x \to 1+}\frac{25(5^{x-1}-1)^2\cdot (\sqrt{x^2-1}+\sqrt{2x-2})}{(x-1)^2}=\left[\begin{array}{ccc}\displaystyle \lim_{x \to 0}\frac{a^x-1}{x}=\ln a\end{array}\right] =\\ \\ \\ =\lim_{x \to 1+}25(\ln 5)^2\cdot (\sqrt{x^2-1}+\sqrt{2x-2})=25\ln^25\cdot 0=0$