Limits at Infinity?

Question

What is the limit as x approaches + infinity of (√4x^2-1) / x^2

My book says that if the degree of numerator and denominator are equal, the limit of the rational function is the ratio of the leading coefficients.
I think the limit is (√4) / 1 = 2 am I correct?

Answer 1

What is under the square root?
Is it sqrt(4)x^2 - 1, where the 4 is the only thing under the root sign? or is it sqrt(4x^2 -1)?
If it's the first, then the answer is sqrt(4)/1 = 2
If its the second, then it is 0 because the degree of the numerator is 1 and the degree of the denominator is 2.

Answer 2

The answer is correct but the solution needs refinement

SQRT((4x^2-1)/x^2)=
= SQRT(4-1/x^2))
Applying the limit of x approaching infinity we have

SQRT(4-0)=2

OR

If we have SQRT(4x^2-1) / x^2 =
SQRT (4/(x^2) - 1/(x^4))
at x-> infinity
=SQRT( 0+0)=0

I hope that helps

Answer 3

if it's sqrt(4x^2 - 1)/(x^2), then the limit as x->inf should be zero because the numerator goes to 2x and the denominator goes to x^2. So then as x->inf, the ratio goes to 2x/(x^2)->2/x->0.

Answer 4

Just put the denominator under the Sqr

Limt { Sqr ( 4x^2-1)} / x^2
x-->00

Limt { Sqr ( 4x^2-1) / x^4 }
x-->00

Limt { Sqr [ ( 4/x^2)-(1/x^4) ] } = 0
x-->00

But if it's in the form

Limt { Sqr ( 4x^2-1) / x^2 }
x-->00

Then
Limt { Sqr [ 4- (1/ x^2)] } = Sqr(4) = 2
x-->00

Answer 5

lim x->+infinity √(4x² - 1)/x²

Divide numerator and denominator by x²
= lim x->+infinity [√(4x² - 1)/x²] / [x²/x²]

Put the numerator x² inside the radical sign by squaring it:
= lim x->+infinity [√[(4x² - 1) / x^4]] / 1

simplify
= lim x->+infinity √(4/x² - 1/x^4)

Use lim x->+infinity 1/x^n = 0
= 0 - 0

= 0

Therefore, the limit is 0

^_^

Answer 6

The denominator's greater degree means it's blowing up much faster than the numerator. The answer is zero.

Answer 7

as long as the square root is only over the four then you are right