Help. Wyznacz granice. Analiza matematyczna

Cześć, wszystko sprowadzasz do wzoru: [latex] lim_{x o infty} (1+ frac{a}{Box} )^{Box} = e^a[/latex] a) [latex]lim_{x o infty} (1+ frac{4}{5x} )^{4x} = lim_{x o infty}[ (1+ frac{4}{5x} )^{5x}}] ^{ frac{4x}{5x} } = (e^4 ) * ^{ frac{4}{5} }= e^{ frac{16}{5} } [/latex] b) [latex] lim_{x o infty} (frac{3x}{1+3x} )^{12x} = lim_{x o infty} (frac{3x+1}{3x} )^{-12x} = lim_{x o infty} (frac{3x+1}{3x} )^{-12x} = \ lim_{x o infty} (1+ frac{1}{3x})^{-12x} = lim_{x o infty} [ (1+ frac{1}{3x})^{3x}] ^{ frac{-12x}{3x} } = (e^1) ^{-4} = frac{1}{e^4} [/latex] c) [latex] lim_{x o infty} ( frac{x+8}{x+12} )^{2x-1 }= lim_{x o infty} ( frac{x+12-12+8}{x+12} )^{2x-1} = \ lim_{x o infty} ( frac{x+12}{x+12} + frac{-4}{x+12} )^{2x-1} = lim_{x o infty}[ ( 1+ frac{-4}{x+12} )^{x+12}] ^{ frac{2x-1}{x+12} } =[e^{-4}]^{2}= \ e^{-8} = frac{1}{e^8} \ [/latex]

Rozwiązanie w załączniku.