lim = (x+5)/(sqrt 4x^2-5) x-->-infinity Solution Take the limit: lim_(x->-infini
ID: 2140455 • Letter: L
Question
lim = (x+5)/(sqrt 4x^2-5)
x-->-infinity
Explanation / Answer
Take the limit: lim_(x->-infinity) (5+x)/(-5+2 sqrt(x^2)) The limit of a sum is the sum of the limits: = 5 (lim_(x->-infinity) 1/(-5+2 sqrt(x^2)))+lim_(x->-infinity) x/(-5+2 sqrt(x^2)) The limit of a quotient is the quotient of the limits: The limit of a constant is the constant: = lim_(x->-infinity) x/(-5+2 sqrt(x^2))+5/(lim_(x->-infinity) (-5+2 sqrt(x^2))) Indeterminate form of type infinity/infinity. Using L'Hospital's rule we have, lim_(x->-infinity) x/(-5+2 sqrt(x^2)) = lim_(x->-infinity) (( dx)/( dx))/(( d(-5+2 sqrt(x^2)))/( dx)): = lim_(x->-infinity) sqrt(x^2)/(2 x)+5/(lim_(x->-infinity) (-5+2 sqrt(x^2))) Factor out constants: = 1/2 (lim_(x->-infinity) sqrt(x^2)/x)+5/(lim_(x->-infinity) (-5+2 sqrt(x^2))) Simplify radicals, sqrt(x^2)/x = -1: The limit of a constant is the constant: = -1/2+5/(lim_(x->-infinity) (-5+2 sqrt(x^2))) The limit of a constant is the constant: The limit of a sum is the sum of the limits: = -1/2+5/(-5+2 (lim_(x->-infinity) sqrt(x^2))) Using the power law, write lim_(x->-infinity) sqrt(x^2) as sqrt(lim_(x->-infinity) x^2): = -1/2+5/(-5+2 sqrt(lim_(x->-infinity) x^2)) Using the power law, write lim_(x->-infinity) x^2 as (lim_(x->-infinity) x)^2: = -1/2+5/(-5+2 sqrt((lim_(x->-infinity) x)^2)) The limit of x as x approaches -infinity is -infinity: Answer: | | = -1/2
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