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Evaluate a. lim_(x→∞) (((x^2 +3)/(27x^2 −1)))^(1/3) b. lim_(x→−∞) ((x−2)/(√(x^2 +1))) c. lim_(x→∞) ((x^2 +2)/(2x−3)) d. lim_(x→∞) 2x + 1 − (√(4x^2 +5))

$${Evaluate} \\ $$$${a}.\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{3}}]{\frac{{x}^{\mathrm{2}} +\mathrm{3}}{\mathrm{27}{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$${b}.\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{{x}−\mathrm{2}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${c}.\underset{{x}\rightarrow\infty} {\:\mathrm{lim}}\frac{{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}{x}−\mathrm{3}} \\ $$$${d}.\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{2}{x}\:+\:\mathrm{1}\:−\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{5}} \\ $$

Question Number 75930 Answers: 0 Comments: 0

consider the function f(x) = { ((1+2x^2 , if x is rational)),((1 + x^4 , if x is irrational)) :} Use the sandwich(pinchin) theorem to prove that lim_(x→0) f(x) = 1.

$${consider}\:{the}\:{function} \\ $$$$\:{f}\left({x}\right)\:=\:\begin{cases}{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} ,\:{if}\:{x}\:{is}\:{rational}}\\{\mathrm{1}\:+\:{x}^{\mathrm{4}} ,\:{if}\:{x}\:{is}\:{irrational}}\end{cases} \\ $$$${Use}\:{the}\:{sandwich}\left({pinchin}\right)\:{theorem}\:{to} \\ $$$${prove}\:{that}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left({x}\right)\:=\:\mathrm{1}. \\ $$

Question Number 75929 Answers: 0 Comments: 2

Evaluate lim_(x→−∞) [(√(1−xe^x )) ]

$${Evaluate} \\ $$$$\underset{{x}\rightarrow−\infty} {\:\mathrm{lim}}\:\left[\sqrt{\mathrm{1}−{xe}^{{x}} }\:\right] \\ $$

Question Number 75925 Answers: 0 Comments: 0

Let OA^(→) = i+3j−2k and OB^(→) =3i+j−2k. The vector OC^(→) bisecting the ∠AOB and C being a point on the line AB is

$$\mathrm{Let}\:\overset{\rightarrow} {{OA}}=\:\boldsymbol{\mathrm{i}}+\mathrm{3}\boldsymbol{\mathrm{j}}−\mathrm{2}\boldsymbol{\mathrm{k}}\:\mathrm{and}\:\overset{\rightarrow} {{OB}}=\mathrm{3}\boldsymbol{\mathrm{i}}+\boldsymbol{\mathrm{j}}−\mathrm{2}\boldsymbol{\mathrm{k}}. \\ $$$$\mathrm{The}\:\mathrm{vector}\:\overset{\rightarrow} {{OC}}\:\mathrm{bisecting}\:\mathrm{the}\:\angle{AOB}\:\mathrm{and} \\ $$$${C}\:\:\mathrm{being}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{line}\:{AB}\:\mathrm{is} \\ $$

Question Number 75924 Answers: 0 Comments: 3

Thr value of ∫_(0 ) ^(π/2) cosec (x−(π/3))cosec (x−(π/6))dx is

$$\mathrm{Thr}\:\mathrm{value}\:\mathrm{of} \\ $$$$\:\underset{\mathrm{0}\:} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{cosec}\:\left({x}−\frac{\pi}{\mathrm{3}}\right)\mathrm{cosec}\:\left({x}−\frac{\pi}{\mathrm{6}}\right){dx}\:\:\mathrm{is} \\ $$

Question Number 75921 Answers: 0 Comments: 1

Question Number 75916 Answers: 0 Comments: 1

x^3 +x^2 −24x+36=0 prove that x=2,3,−6.

$${x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{36}=\mathrm{0} \\ $$$${prove}\:{that}\:{x}=\mathrm{2},\mathrm{3},−\mathrm{6}. \\ $$

Question Number 75913 Answers: 0 Comments: 1

x^3 −7x+6=0 prove that x=2,−3,1 .

$${x}^{\mathrm{3}} −\mathrm{7}{x}+\mathrm{6}=\mathrm{0} \\ $$$${prove}\:{that}\:{x}=\mathrm{2},−\mathrm{3},\mathrm{1}\:. \\ $$

Question Number 75918 Answers: 1 Comments: 0

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Question Number 75902 Answers: 1 Comments: 1

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Question Number 75899 Answers: 0 Comments: 0

A group of 30 men participate in a survey of telecom companies. The number of men who use both Idea and Airtel internet was equal to the number of men who use neither of these inernets. The number of men who use Idea is 4 more than those who use Airtel. How many use Airtel internet.

$$\mathrm{A}\:\mathrm{group}\:\mathrm{of}\:\mathrm{30}\:\mathrm{men}\:\mathrm{participate}\:\mathrm{in}\:\mathrm{a}\: \\ $$$$\mathrm{survey}\:\mathrm{of}\:\mathrm{telecom}\:\mathrm{companies}.\:\mathrm{The} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{men}\:\mathrm{who}\:\mathrm{use}\:\mathrm{both}\:\mathrm{Idea}\: \\ $$$$\mathrm{and}\:\mathrm{Airtel}\:\mathrm{internet}\:\mathrm{was}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{men}\:\mathrm{who}\:\mathrm{use}\:\mathrm{neither} \\ $$$$\mathrm{of}\:\mathrm{these}\:\mathrm{inernets}.\:\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{men} \\ $$$$\mathrm{who}\:\mathrm{use}\:\mathrm{Idea}\:\mathrm{is}\:\mathrm{4}\:\mathrm{more}\:\mathrm{than}\:\mathrm{those} \\ $$$$\mathrm{who}\:\mathrm{use}\:\mathrm{Airtel}.\:\mathrm{How}\:\mathrm{many}\:\mathrm{use}\:\mathrm{Airtel} \\ $$$$\mathrm{internet}. \\ $$

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