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
VNVDVI
Mar 1, 2018

#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.

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