How do you determine the limit of #(x+5)/(x-2)# as x approaches 2+?

1 Answer
Mar 3, 2018

The limit tends to #oo#.

Explanation:

We have to solve #lim_(xrarr2^+)(x+5)/(x-2)#

Directly inputting will not work, as the denominator will give #0#, and division by #0# is not possible.

So we make #x# get closer and closer to #2#.
For example, we take #2.1#, then #2.01#, #2.001#, and so on.

For #2.1#:

#(2.1+5)/(2.1-2)=7.1/0.1=71#

For #2.01#:

#(2.01+5)/(2.01-2)=7.01/0.01=701#

And so on.

We see a pattern. The closer we get to #x=2#, the value of the limit increases dramatically.

This is because the division of a number by an extremely small one gives us an extremely large answer.

Therefore, we can say:

#lim_(xrarr2)(x+5)/(x-2)=+-oo#

However, the questions asks us to get to two from values larger than it, as shown in the examples. This gives us a positive denominator and numerator, showing us that, when #xrarr2^+#, #-oo# is not valid.

So we say #lim_(xrarr2^+)(x+5)/(x-2)=oo#