Lim x-> infinity 7/((e^x)+2) evaluate ?

1 Answer
Anthony R.
Jan 19, 2018

#lim_(x->oo)7/(e^x+2)=0#

Explanation:

Let try to substitute #oo# for #x#

Recall: #lim_(x->oo)e^x=oo#

This can easily be verified by taking a look of the graph of #e^x# (See Below)

graph{e^x [-9.58, 10.42, -1.24, 8.76]}

#lim_(x->oo)7/(e^x+2)=7/(e^color(red)oo+2)=7/(oo+2)=7/oo=0#

As to why #7/oo# tends to #0# think of it like this:

As #x# approaches infinity the denominator gets larger and larger towards infinity but the fraction as a whole gets smaller and smaller. Here's a few examples:

Consider:

#1/10=0.1#

#1/100=0.01#

#1/1000=0.001#

#1/10000000=0.0000001#

As you can see, the fraction becomes smaller and smaller/ closer and closer to #0#. The fact that the numerator is #7# instead of #1# doesn't change this fact.

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