Question

How do you determine the limit of `sqrt(x-4)/(3x+5)` as x approaches negative infinity?

Answer 1

`1/3`

Explanation:

Notice that you have a square root in your numerator. This can complicate the usual process, so you'll need to take a few extra steps to make sure this works.

The first step is the same as usual: we're going to divide everything by the highest power, which in this case is `x^1`:

`lim_(x->-oo)(1/x)/(1/x) * sqrt(x-4)/(3x + 5)`

For the numerator, we have to make one small simplification to get the right number into the square root:

`=> lim_(x->-oo)(1/sqrt(x^2))/(1/x) * sqrt(x-4)/(3x + 5)`

`=> lim_(x->-oo) sqrt((x-4)/x)/((3x + 5)/x)`

Now, we divide through and simplify:

`=> lim_(x->-oo) sqrt((1 - 4/x))/(3 + 5/x)`

Now, you just go ahead and take your limit:

`=> sqrt((1 - cancel(4/x)))/(3 + cancel(5/x))`

`=>1/3`

Hope that helped :)