What is the limit of (25*x^2+5*x+16)^(1/2)-5*x+16 as x goes to infinity?

Question

Square root of (25x^2 + 5x +16) -5x +16

Answer 1

NOTE: Infinity minus infinity does not necessarily approach 0.

In cases like these, you rationalize the numerator.

[sqrt(25x^2+5x+16) - (5x-16)]*
[sqrt(25x^2+5x+16) + (5x-16)]/[sqrt(25x^2+5x+16) + (5x-16)]

=[(25x^2+5x+16) - (5x-16)^2]/[sqrt(25x^2+5x+16) + (5x-16)]
=(165x-240)/[sqrt(25x^2+5x+16) + (5x-16)]

Divide numerator and denomator by x

=(165-240/x)/[sqrt(25+5/x+16/x^2) + (5-16/x)]

Now take the limit as x approaches infinity

Lim = (165-0)/[sqrt(25+0+0) + (5-0)]
= 165/(10)
= 33/2

Answer 2

16

If we substitute x = infinity into the function Sqrt(25x^2 + 5^x + 16) - 5x + 16, we get:

sqrt(infinity) - 5(infinity) + 16.

In theory, sqrt(infinity) and 5(infinty) are both infinity. Thus, we have

infinity - infinity + 16

which equals 0+16.

Therefore, the limit equals 16.

Answer 3

as x -> oo sqrt(25*x^2) = 5x
so the eq -> 5x - 5x + 16 = 16

Answer 4

x -> pos infinity ( +oo)

f(x) = (25*x^2+5*x+16)^(1/2)-5*x+16
=( (25*x^2+5*x+16)^(1/2)-5*x+16 ) * ((25*x^2+5*x+16)^(1/2)+ 5*x- 16 )/ ((25*x^2+5*x+16)^(1/2)+ 5*x- 16 )
=( 25*x^2+5*x+16 -25x^2+ 160x - 196)/
((25*x^2+5*x+16)^(1/2)+ 5*x- 16 )
= (165x- 180)/((25*x^2+5*x+16)^(1/2)+ 5*x- 16 )

lim f = lim (165x-180)/((25*x^2+5*x+16)^(1/2)+ 5*x- 16 )
lim( 165-180/x)/(( 25+5/x+16/x^2)^1/2+ 5+16/x)
= 165/(5+5)
=33/2

x -> neg infinity ( -oo)

(25*x^2+5*x+16)^(1/2) -> +oo
-5*x+16 -> +oo
=> f -> +oo