Question

What is the limit of `(x +sqrt((x^2)+(3x)))` as x goes to infinity?

Answer 1

`lim_(xrarroo) (x +sqrt(x^2+3x)) = oo`

Explanation:

`(x +sqrt(x^2+3x)) = x +sqrt(x^2(1+3/x))` for `x != 0`

`= x +sqrt(x^2)sqrt(1+3/x))` for `x != 0`

`= x +absx sqrt(1+3/x))` for `x != 0`

Now, as `xrarroo`, we have

`absx rarroo` and

`sqrt(1+3/x) rarr sqrt(1+0) = 1`

So,

`lim_(xrarroo) (x +sqrt(x^2+3x)) = lim_(xrarroo) (x +absx sqrt(1+3/x)) = oo`

Bonus

As `x` decreases without bound, the limit is quite different.

As `x rarr -oo`, we get indeterminate form: `-oo+oo`.

We think of `(x +sqrt(x^2+3x))` as a ratio (over `1`) and eliminate the square root from the numerator.

`(x +sqrt(x^2+3x)) * (x -sqrt(x^2+3x)) /(x -sqrt(x^2+3x)) = (-3x)/(x -sqrt(x^2+3x))`

`= (-3x)/(x-absx sqrt(1+3/x))`.

For `x < 0`, we have `absx = -x`, so we have:

`lim_(xrarr-oo)(x +sqrt(x^2+3x)) = lim_(xrarr-oo)(-3x)/(x-absx sqrt(1+3/x))`

`= lim_(xrarr-oo)(-3x)/(x+xsqrt(1+3/x))`

`= lim_(xrarr-oo)(-3)/(1+sqrt(1+3/x))`

`= -3/2`.

Answer 2

`lim_(x->+oo) (x+sqrt(x^2+3x)) = +oo`
`lim_(x->-oo) (x+sqrt(x^2+3x)) = -3/2`

Explanation:

`sqrt(x^2+3x) >= 0` for all `x >= 0`

So

`lim_(x->+oo) (x+sqrt(x^2+3x)) >= lim_(x->+oo) x = +oo`

Much more interesting is the case `lim_(x->-oo) (x+sqrt(x^2+3x))`

First notice that `lim_(t->oo) (t - sqrt(t^2-C)) = 0` for any constant `C > 0`

To see this: Given `epsilon > 0`

`(t-epsilon)^2 = t^2 - 2tepsilon + epsilon^2`

So if `t > (C+epsilon^2)/(2epsilon)`, then

`(t-epsilon)^2 = t^2 - 2tepsilon + epsilon^2 < t^2 - (C +epsilon^2) + epsilon^2 = t^2 - C`

So

`t-epsilon < sqrt(t^2-C) < sqrt(t^2) = t`

Let `t = -(x+3/2)` and `C = 9/4`

Then:

`0 = lim_(t->oo) (sqrt(t^2-C) - t)`

`= lim_(x->-oo) (sqrt((x+3/2)^2 - 9/4) + x + 3/2)`

`= lim_(x->-oo) (sqrt(x^2+3x+9/4 - 9/4) + x + 3/2)`

`= lim_(x->-oo) (sqrt(x^2+3x) + x + 3/2)`

`= lim_(x->-oo) (sqrt(x^2+3x) + x) + 3/2`

Hence:

`lim_(x->-oo) (sqrt(x^2+3x) + x) = -3/2`