Question

$lim_{x to 0} e^4^x-e^-^2^x/2arctgx-sinx$

Answer 1

$lim_{x to 0} e^{4x} - e^{ frac{-2x}{2arctg(x) - sin(x)} } lim_{x to 0}e^{4x} - lim_{x to 0}e^{ frac{-2x}{2arctg(x) - sin(x)} } 1 - lim_{x to 0}e^{ frac{-2x}{2arctg(x) - sin(x)} } 1 - e^{lim_{x to 0} frac{-2x}{2arctg(x) - sin(x)} } 1 - e^{ frac{lim_{x to 0}-2x}{lim_{x to 0}(2arctg(x) - sin(x))} } 1 - e^{ -2frac{-x}{lim_{x to 0}(2arctg(x) - lim_{x to 0sin(x))} } 1 - e^{ -2frac{-x}{lim_{x to 0}(2arctg(x) - x} } 1 - e^{ -2frac{-x}{0 - x} 1 - e^{ -2}$


$lim_{x to 0} e^{4x} - e^{ frac{-2x}{2arctg(x) - sin(x)} } = 1 -e^{-2}$

Answer 2

нет так ,2arctgx -sin x -это знаменатель дроби

Answer 3

не в степени(