Answer to Question #246054 in Calculus for Annie

Question #246054

Question No#08

Find the limit of each rational function as 𝒙 β†’ ±∞. 𝑓(π‘₯) = 3π‘₯ + 7

Β π‘₯2 βˆ’ 2

𝑓(π‘₯)= 10π‘₯5 +π‘₯4 +31. π‘₯6


Question No#10

Let

Ζ’(x) = π‘₯1/(1βˆ’x) .

Make tables of values of Ζ’ at values of x that approach x = 1 from above and below. Does Ζ’(x)

appear to have a limit as x approaches 1? If so, what is it? If not, why not?

Β Question No#11

Find the limits using π₯𝐒𝐦 𝐬𝐒𝐧 𝜽 = 𝟏 π’™β†’πŸŽ 𝜽

a) lim 6π‘₯2(cot π‘₯)(csc 2π‘₯) π‘₯β†’0

𝑏) lim sin 3𝑦 cot 5𝑦 𝑦→0 𝑦 cot4𝑦


1
Expert's answer
2021-10-06T17:23:17-0400

Question No#08: The first two limits are both equal to zero:


"\\lim\\limits_{x\\, \\to \\,\u00b1\\infin} \\cfrac{3x+7}{x^2-2}=\\lim\\limits_{x\\, \\to \\,\u00b1\\infin} \\cfrac{3x+7}{x^2-2} \\Bigg( \\dfrac{\\frac{1}{x^2}}{\\frac{1}{x^2}} \\Bigg) = \n\\\\ =\\lim\\limits_{x\\, \\to \\,\u00b1\\infin} \\cfrac{\\frac{3}{x}+\\frac{7}{x^2}}{1-\\frac{2}{x^2}}= \\cfrac{0+0}{1-0}=\\cfrac{0}{1}=0"


"\\lim\\limits_{x\\, \\to \\,\u00b1\\infin} \\cfrac{10x^5+x^4 +31}{x^6}\n\\\\ =\\lim\\limits_{x\\, \\to \\,\u00b1\\infin} \\cfrac{10x^5+x^4 +31}{x^6} \\Bigg( \\dfrac{\\frac{1}{x^6}}{\\frac{1}{x^6}} \\Bigg) \n\\\\ =\\lim\\limits_{x\\, \\to \\,\u00b1\\infin} \\cfrac{\\frac{10}{x}+\\frac{1}{x^2}+\\frac{31}{x^6}}{1}= \\cfrac{0+0+0}{1}=0"



Question No#10

Let Ζ’(x) = π‘₯1/(1βˆ’x) . Make tables of values of Ζ’ at values of x that approach x = 1 from above and below.




Does Ζ’(x) appear to have a limit as x approaches 1? Yes, there is a limit.

If so, what is it? The limit for "f(x)=\\cfrac{x}{1-x}" can be expressed as:"\\lim\\limits_{x\\, \\to \\,1^+} f(x)=+\\infin \\text{ and} \\lim\\limits_{x\\, \\to \\,1^-} f(x)=-\\infin"


Question No#11


Find the limits using "\\lim\\limits_{\\theta\\, \\to \\,0} \\cfrac{\\sin\\theta}{\\theta}=1", after substitution we find


a) "\\lim\\limits_{x\\, \\to \\,0} 6x^2(\\cot x)(\\csc 2x)\n\\\\ =\\lim\\limits_{x\\, \\to \\,0} 6x^2(\\frac{\\cos x}{\\sin x})(\\frac{1}{2\\sin x \\cos x})\n\\\\ =\\lim\\limits_{x\\, \\to \\,0} (\\frac{3x^2}{\\sin^2 x})=3(\\lim\\limits_{x\\, \\to \\,0} (\\frac{x}{\\sin x}))^2\n\\\\ \\therefore \\lim\\limits_{x\\, \\to \\,0} 6x^2(\\cot x)(\\csc 2x) =3(1^2)=3"


𝑏) "\\lim\\limits_{y\\, \\to \\,0} \\cfrac{(\\sin^3y)(\\cot^ 3y)}{y\\cot^ 4y}\n\\\\ =\\lim\\limits_{y\\, \\to \\,0} \\cfrac{\\sin^3y}{y\\cot y} =\\lim\\limits_{y\\, \\to \\,0} \\cfrac{\\sin^3y \\cos y}{y\\sin y} \n\\\\ =\\lim\\limits_{y\\, \\to \\,0} \\cfrac{\\sin^2y \\cos y}{y}\n\\\\ =\\lim\\limits_{y\\, \\to \\,0} \\cfrac{\\sin y }{y} \\times \\lim\\limits_{y\\, \\to \\,0} {\\sin y \\cos y}\n\\\\ =(1)(0)=0"


Reference:

  • Thomas, G. B., & Finney, R. L. (1961).Β Calculus. Addison-Wesley Publishing Company.

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