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

1 Answer
Mar 1, 2018

11

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.