Answer to Question #213706 in Calculus for rahul

Question #213706

Evaluate the following limits

(a) Lim y"\\to" -1 Fourth root of 4y^3 minus 3

(b) lim x"\\to" infinity 1+x all over 1-x

(c) lim x "\\to" 2 x - 2 all over 4 - x^2

(d) lim x "\\to" 0 sinx - x cos x all over x^3


1
Expert's answer
2021-07-26T17:31:37-0400

(a) "\\lim\\limits_{y\\to-1}\\sqrt[4]{4y^3-3}=?"

If "y<0" then "4y^3-3<0" and the fourth root doesn't exist in real numbers.

So we will consider another limit:

"\\lim\\limits_{y\\to 1}\\sqrt[4]{4y^3-3}=\\lim\\limits_{y\\to 1}\\sqrt[4]{1+4(y^3-1)}=\\lim\\limits_{y\\to 1}\\sqrt[4]{1}=1"


(b) "\\lim\\limits_{x\\to\\infty}\\frac{1+x}{1-x}=\\lim\\limits_{x\\to\\infty}\\frac{x^{-1}+1}{x^{-1}-1}=\\lim\\limits_{x\\to\\infty}\\frac{1}{-1}=-1"


(c) "\\lim\\limits_{x\\to 2}\\frac{x-2}{4-x^2}=\\lim\\limits_{x\\to 2}\\frac{x-2}{(2+x)(2-x)}=\\lim\\limits_{x\\to 2}\\frac{-1}{x+2}=-\\frac{1}{4}"


(d) "\\lim\\limits_{x\\to 0}\\frac{\\sin x-x\\cos x}{x^3}=\\lim\\limits_{x\\to 0}\\frac{x-x^3\/6+o(x^3)-x(1-x^2\/2+o(x^2))}{x^3}="

"=\\lim\\limits_{x\\to 0}\\frac{-x^3\/6+x^3\/2+o(x^3)}{x^3}=\\lim\\limits_{x\\to 0}(-\\frac{1}{6}+\\frac{1}{2}+o(1))=\\frac{1}{3}"


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