Question

I was asked to evaluate the following limit expression: `lim_(xtooo)(3x-2)/(8x+7)` Please show all steps. ? Thanks

Answer 1

`lim_(xrarroo) [(3x-2)/(8x+7)] = color(blue)(3/8`

Explanation:

Here are two different methods you may use for this problem different than Douglas K. 's method of using l'Hôpital's rule.

We're asked to find the limit

`lim_(xrarroo) [(3x-2)/(8x+7)]`

The simplest way you can do this is plug in a very large number for `x` (such as `10^10`) and see the outcome; the value that comes out is generally the limit (you may not always do this, so this method is usually ill-advised):

`(3(10^10)-2)/(8(10^10)+7) ~~ color(blue)(3/8`

However, the following is a surefire way to find the limit:

We have:

`lim_(xrarroo) [(3x-2)/(8x+7)]`

Let's divide the numerator and denominator by `x` (the leading term):

`lim_(xrarroo) [(3-2/x)/(8+7/x)]`

Now, as `x` approaches infinity, the values `-2/x` and `7/x` both approach `0`, so we're left with

`lim_(xrarroo) [(3-(0))/(8+(0))] = color(blue)(3/8`

Answer 2

Because the expression evaluated at the limit is the indeterminate form `oo/oo`, the use of L'Hôpital's rule is warranted.

Explanation:

Use L'Hôpital's rule:

`Lim_(xtooo)(d((3x-2))/dx)/((d(8x+7))/dx) =`

`Lim_(xtooo)3/8 = 3/8`

The rule says that the limit of the original expression is the same:

`Lim_(xtooo)(3x-2)/(8x+7) = 3/8`