Question

What is the limit of `sqrt(4x^2-1) / x^2` as x goes to negative infinity?

Answer 1

The limit is zero.

Explanation:

As `x` approaches `-\infty`, the `-1` inside the root becomes insignificant (which is to say, `4x^2` and `4x^2-1` are asymptotically equivalent, which again means that you can substitute one with the other and the limits doesn't change).

So, you have the following:

`\lim_{x\to -\infty} {\sqrt{4x^2-1}}/{x^2} = \lim_{x\to -\infty} {\sqrt{4x^2}}/{x^2}`

With the advantage that `\sqrt{4x^2}` is easibly computable, since it is `abs(2x)`. And since `{abs(2x)}/x^2=2/abs(x)`, you can clearly see that this function goes to `0`, as `x` approaches `-\infty`.

Answer 2

Here is a different way of writing the solution.

Explanation:

`sqrt(4x^2-1) = sqrt(x^2(4-1/x^2))`

`= sqrt(x^2)sqrt(4-1/x^2)`

`= absx sqrt(4-1/x^2)`

As `x rarr-oo` we are concerned with negative values of `x`,.

For `x < 0`, `sqrt(x^2) = absx = -x`.

`lim_(xrarr-oo)sqrt(4x^2-1)/x^2 = lim_(xrarr-oo)(sqrt(x^2)sqrt(4-1/x^2))/x^2`

`= lim_(xrarr-oo)(-xsqrt(4-1/x^2))/x^2`

`= lim_(xrarr-oo)(-sqrt(4-1/x^2))/x`

`= 0`.

(I would describe this as more algebraic and less analytic than the other fine answer provided by Killer Bunny (and George C.).)