Question No#08
Find the limit of each rational function as π β Β±β. π(π₯) = 3π₯ + 7
Β π₯2 β 2
π(π₯)= 10π₯5 +π₯4 +31. π₯6
Question No#10
Let
Ζ(x) = π₯1/(1βx) .
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₯π’π¦ π¬π’π§ π½ = π πβπ π½
a) lim 6π₯2(cot π₯)(csc 2π₯) π₯β0
π) lim sin 3π¦ cot 5π¦ π¦β0 π¦ cot4π¦
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/(1βx) . 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"
π) "\\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:
Comments
Leave a comment