Ответ:

[tex] \frac{lim}{x - > 0} ( \frac{5x}{arcsin(15x)} )[/tex]

[tex] \frac{lim}{x - > 0} (5x) \\ \frac{lim}{x - > 0} (arcsin(15x))[/tex]

[tex]0 \\ 0[/tex]

[tex] \frac{lim}{x - > 0} ( \frac{5x}{arcsin(15x)} )[/tex]

правило Лопиталя:

[tex] \frac{lim}{x - > c} ( \frac{f(x)}{g(x)} ) = \frac{lim}{x - > c} ( \frac{f'(x)}{g'(x)} )[/tex]

[tex] \frac{lim}{x - > 0} ( \frac{ \frac{d}{dx} (5x)}{ \frac{d}{dx}(arcsin(15x)) } )[/tex]

[tex] \frac{d}{dx}(5x) = 5[/tex]

[tex] \frac{d}{dx} (arcsin(15x)) = \frac{d}{dg} (arcsin(g)) \times \frac{d}{dx} (15x) = \frac{1}{ \sqrt{1 - {g}^{2} } } \times 15 = \frac{1}{ \sqrt{1 - {(15x)}^{2} } } \times 15 = \frac{15}{ \sqrt{1 - {225x}^{2} } } [/tex]

[tex] \frac{lim}{x - > 0} ( \frac{5}{ \frac{15}{ \sqrt{1 - {225x}^{2} } } } )[/tex]

[tex] \frac{lim}{x - > 0} (5 \div \frac{15}{ \sqrt{1 - {225x}^{2} } } )[/tex]

[tex] \frac{lim}{x - > 0} (5 \times \frac{ \sqrt{1 - {225x}^{2} } }{15} )[/tex]

[tex] \frac{lim}{x - > 0} ( \frac{5 \sqrt{1 - {225x}^{2} } }{15} )[/tex]

сокращаем на общий делитель 5:

[tex] \frac{lim}{x - > 0} ( \frac{ \sqrt{1 - {225x}^{2} } }{3} )[/tex]

[tex] \frac{ \frac{lim}{x - > 0}( \sqrt{1 - {225x}^{2} }) }{ \frac{lim}{x - > 0} (3)} [/tex]

[tex] \frac{ \sqrt{ \frac{lim}{x - > 0}(1 - {225x}^{2}) } }{3} [/tex]

[tex] \frac{ \sqrt{ \frac{lim}{x - > 0} (1) - \frac{lim}{x - > 0}( {225x}^{2} )} }{3} [/tex]

[tex] \frac{ \sqrt{1 - 225 \times \frac{lim}{x - > 0}( {x}^{2}) } }{3} [/tex]

[tex] \frac{ \sqrt{1 - 225 \times ( \frac{lim}{x - > 0} (x)) ^{2} } }{3} [/tex]

[tex] \frac{ \sqrt{1 - 225 \times {0}^{2} } }{3} [/tex]

[tex] \frac{ \sqrt{1 - 225 \times 0} }{3} [/tex]

[tex] \frac{ \sqrt{1 - 0} }{3} [/tex]

[tex] \frac{ \sqrt{1} }{3} [/tex]

[tex] \frac{1}{3} [/tex]