Question

What is the limit? `lim_(x to -oo) (x+2)/(sqrt(9x^2+1)`

Answer 1

`lim_(x to -oo) (x+2)/(sqrt(9x^2+1)) = -1/3`

Explanation:

Given: `lim_(x to -oo) (x+2)/(sqrt(9x^2+1))`

Bring the numerator under the radical but the radical will be negative for large negative values of x:

`lim_(x to -oo) -sqrt((x+2)^2/(9x^2+1))`

Expand the square:

`lim_(x to -oo) -sqrt((x^2+4x+4)/(9x^2+1))`

Factor out `1/3`:

`lim_(x to -oo) -1/3sqrt((x^2+4x+4)/(x^2+1/9))`

Separate 4 into `1/9+35/9`:

`lim_(x to -oo) -1/3sqrt((x^2+1/9+4x+35/9)/(x^2+1/9))`

Separate into two fractions:

`lim_(x to -oo) -1/3sqrt((x^2+1/9)/(x^2+1/9)+(4x+35/9)/(x^2+1/9))`

The first fraction becomes 1:

`lim_(x to -oo) -1/3sqrt(1+(4x+35/9)/(x^2+1/9))`

The limit is now on the second term:

`-1/3sqrt(1+lim_(x to -oo)(4x+35/9)/(x^2+1/9))`

The limit evaluates to `(-oo)/oo`, therefore, we use L'Hôpital's rule :

`-1/3sqrt(1+lim_(x to -oo)((d(4x+35/9))/dx)/((d(x^2+1/9))/dx))`

Evaluate the derivatives:

`-1/3sqrt(1+lim_(x to -oo)2/x)`

We know this limit to be 0:

`-1/3sqrt(1+0) = -1/3`

Answer 2

`-1/3`

Explanation:

`lim_{x to -infty} {x+2}/sqrt{9x^2+1} = lim_{x to -infty} {x+2}/{3|x|sqrt{1+1/{9x^2}}}`
# = lim_{x to -infty} x/{3|x|} {1+2/x}/{sqrt{1+1/{9x^2}}} = lim_{x to -infty} x/{3|x|} lim_{x to -infty} {1+2/x}/{sqrt{1+1/{9x^2}}} = -1/3 \times 1 = -1/3 #

The trick here is to realize that the square root in calculus always stands for the positive square root and hence `sqrt{9x^2} = 3|x|`