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

2 Answers
Feb 8, 2018

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

Feb 8, 2018

-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|