Question

Limit as x approaches infinity (x+2)/(x+3)?

without using the comparison of degrees of the numerator and denominator. show work in great detail

Answer 1

`1`

Explanation:

Let's attempt to plug in `oo` right away:

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

This indeterminate form indicates that we must somehow simplify. In the case of limits to `+-oo` of rational functions, we can divide both the numerator and denominator by the term of highest exponent in the denominator.

Here, the term of highest exponent in the denominator is `x^1,` or `x.` Let's divide everything, numerator and denominator, by `x:`

`lim_(x->oo)((x+2)/x)/((x+3)/x)=lim_(x->oo)(x/x+2/x)/(x/x+3/x)`
(Because `(a+b)/c=a/c+b/c`)

Simplify:

`lim_(x->oo)(x/x+2/x)/(x/x+3/x)=lim_(x->oo)(1+2/x)/(1+3/x)`

Now, let's plug in `oo:`

`lim_(x->oo)(1+2/x)/(1+3/x)=(1+2/oo)/(1+3/oo)=(1+0)/(1+0)=1`

Because `c/oo,` where `c` is a constant, is `0.` Dividing a constant value by an extremely large value yields a final value of `0,` in other words.