lim x rightarrow infinity (x - x2 + x). Solution answer is -1/2 Take the limit:

Question

Explanation / Answer

answer is -1/2 Take the limit: lim_(x->infinity) (x-sqrt(x^2+x)) Rationalize x-sqrt(x^2+x) = ((x-sqrt(x^2+x)) (sqrt(x^2+x)+x))/(sqrt(x^2+x)+x) = -x/(sqrt(x^2+x)+x): = lim_(x->infinity) -x/(sqrt(x^2+x)+x) Factor out constants: = -(lim_(x->infinity) x/(x+sqrt(x^2+x))) Writex/(sqrt(x^2+x)+x) as 1/(sqrt(x^2+x)/x+1): = -(lim_(x->infinity) 1/(sqrt(x^2+x)/x+1)) The limit of a quotient is the quotient of the limits: The limit of a constant is the constant: = -1/(lim_(x->infinity) (sqrt(x^2+x)/x+1)) The limit of a constant is the constant: The limit of a sum is the sum of the limits: = -1/(lim_(x->infinity) sqrt(x^2+x)/x+1) Simplify radicals, sqrt(x^2+x)/x = sqrt((x^2+x)/x^2): = -1/(lim_(x->infinity) sqrt((x^2+x)/x^2)+1) Using the power law, write lim_(x->infinity) sqrt((x^2+x)/x^2) as sqrt(lim_(x->infinity) (x^2+x)/x^2): = -1/(sqrt(lim_(x->infinity) (x^2+x)/x^2)+1) Indeterminate form of type infinity/infinity. Using L'Hospital's rule we have, lim_(x->infinity) (x^2+x)/x^2 = lim_(x>infinity) (( d(x^2+x))/( dx))/(( dx^2)/( dx)): = -1/(sqrt(lim_(x->infinity) (1+1/(2 x)))+1) The limit of a constant is the constant: The limit of a sum is the sum of the limits: = -1/(sqrt(1/2 (lim_(x->infinity) 1/x)+1)+1) The limit of a quotient is the quotient of the limits: The limit of a constant is the constant: = -1/(sqrt(1/(2 (lim_(x->infinity) x))+1)+1) The limit of x as x approaches infinity is infinity: Answer: | | = -1/2

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.