Answer to Question #107072 in Real Analysis for Garima Ahlawat

Question #107072
(1) check whether the function f given by f(x) = 1/(2x-4)^2 for all x belonging to ]-2,2[ is continuous in the interval ]-2,2[. Is it bounded? Justify your answer.
(2) find the values of p and q so that lim { p sinx + x(1-cosx)}/5x^3 as x tends to infinity is 1/6
1
Expert's answer
2020-03-31T11:02:41-0400

(1) Consider a function "f(x)=\\frac{1}{(2x-4)^2}, \\ x \\in ]-2,2[" . The gived function is continuous on interval "]-2,2[" as a ratio of two continuous functions on "]-2,2[".

The function "f(x)" is unbounded in the interval "]-2,2[" , because there exists a sequence "x_n=2-\\frac1n, \\ n \\in \\mathbb{N}", in interval "]-2,2[" , such that "f(x_n)=\\frac{n^2}{4} \\to \\infty" for "n \\to \\infty".


(2) Let consider a function "g(x)=\\frac{p\\sin x+x(1-\\cos x)}{5x^3}=\\frac{p\\sin x}{5x^3}+\\frac{1-\\cos x}{5x^2}." Since "p\\sin x" and "1-\\cos x" are bounded functions in "\\mathbb{R}", then "\\lim\\limits_{x \\to \\infty}\\frac{p\\sin x}{5x^3}=0", "\\lim\\limits_{x \\to \\infty} \\frac{1-\\cos x}{5x^2}=0". Therefore "\\lim\\limits_{x \\to \\infty} \\frac{p\\sin x+x(1-\\cos x)}{5x^3}=0" for every values of "p" and "q". Consequently, there is no values of parameters "p" and "q" for which "\\lim\\limits_{x \\to \\infty} \\frac{p\\sin x+x(1-\\cos x)}{5x^3}=\\frac{1}{6}."


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