Método de Cambio de Variable II

$d left parenthesis x right parenthesis equals fraction numerator fourth root of x squared minus 3 end root minus 1 over denominator square root of x squared minus 3 end root minus 1 end fraction semicolon space u to the power of 4 equals x squared minus 3 space C u a n d o space x rightwards arrow 2 space e n t o n c e s space u rightwards arrow 1 stack l i m with x rightwards arrow 2 below space fraction numerator fourth root of x squared minus 3 end root minus 1 over denominator square root of x squared minus 3 end root minus 1 end fraction equals space stack l i m with u rightwards arrow 1 below space fraction numerator fourth root of u to the power of 4 end root minus 1 over denominator square root of u to the power of 4 end root minus 1 end fraction equals space stack l i m with u rightwards arrow 1 below space fraction numerator u minus 1 over denominator u squared minus 1 end fraction equals stack l i m with u rightwards arrow 1 below space fraction numerator u minus 1 over denominator open parentheses u minus 1 close parentheses open parentheses u plus 1 close parentheses end fraction stack l i m with u rightwards arrow 1 below space fraction numerator u minus 1 over denominator open parentheses u minus 1 close parentheses open parentheses u plus 1 close parentheses end fraction equals stack l i m with u rightwards arrow 1 below space fraction numerator 1 over denominator open parentheses u plus 1 close parentheses end fraction equals stack l i m with u rightwards arrow 1 below space fraction numerator 1 over denominator open parentheses 1 plus 1 close parentheses end fraction equals 1 half g left parenthesis x right parenthesis equals fraction numerator straight pi minus straight x over denominator sin left parenthesis x right parenthesis end fraction space semicolon space u equals straight pi minus straight x space space space comma space space straight x equals straight pi minus straight u space semicolon space sin left parenthesis straight pi right parenthesis equals 0 semicolon cos left parenthesis straight pi right parenthesis equals negative 1 semicolon space space stack lim space with straight u rightwards arrow 0 below space fraction numerator sin left parenthesis straight u right parenthesis over denominator straight u end fraction equals 1 stack l i m with x rightwards arrow straight pi below space fraction numerator straight pi minus straight x over denominator s i n left parenthesis x right parenthesis end fraction equals stack l i m with u rightwards arrow 0 below space fraction numerator straight u over denominator s i n left parenthesis straight pi minus straight u right parenthesis end fraction equals stack l i m with u rightwards arrow 0 below space fraction numerator straight u over denominator s i n left parenthesis straight pi right parenthesis cos left parenthesis u right parenthesis minus sin left parenthesis u right parenthesis cos left parenthesis straight pi right parenthesis end fraction equals stack l i m with u rightwards arrow 0 below space fraction numerator straight u over denominator negative s i n left parenthesis u right parenthesis open parentheses negative 1 close parentheses end fraction equals stack l i m with u rightwards arrow 0 below space fraction numerator straight u over denominator s i n left parenthesis u right parenthesis end fraction equals stack l i m with u rightwards arrow 0 below space fraction numerator 1 over denominator begin display style fraction numerator sin left parenthesis u right parenthesis over denominator u end fraction end style end fraction equals 1$

En estos ejemplos se puede apreciar el alcance del método de cambio de variable y su ventaja con respecto a otros métodos estudiados. Hay un límite que no se resolvió y se asumió que su solución era uno; este límite es :

$stack l i m space with straight u rightwards arrow 0 below space fraction numerator s i n left parenthesis straight u right parenthesis over denominator straight u end fraction equals 1$

La resolución de este límite escapa al objetivo del curso. De todas formas será abordado en próximas lecciones, posiblemente casi al final del curso, por la complejidad de su resolución.