How do you find #lim (x^2+4)/(x^2-4)# as #x->2^+#?

2 Answers
Feb 10, 2017

One method is to evaluate at values closer and closer to 2.

Explanation:

f(2.1) = 20.5
f(2.05) = 40.5
f(2.01) = 200.5
f(2.0001) = 20000.5

So we can see it approaches infinity.

Another method is by graphing. As you can see, the limit approaches infinity.

graph{(x^2+4)/(x^2-4) [-5.74, 12.04, -1.71, 7.18]}

As we can see as x approaches two from the positive direction, the y value seems to go up indefinitely.

Feb 10, 2017

Please see below.

Explanation:

As #xrarr2#, the numerator goes to #8# (which is not #0#) and the denominator goes to #0#.

This form #("non-"0)/0# tells us that the function is increasing or decreasing without bound on each side. (The one-sided limits are #+-oo#)

To see which is happening as #xrarr2^+#, we need to determine whether the denominator is going to #0# through positive or negative values.

For #x# a little greater than #2#, we know that #x^2# is a little greater than #4#. Therefore, the denominator is positive.

Something close to #8# divided by a positive number close to #0# (a positive fraction) is a big positive number.

#lim_(xrarr2^+)(x^2+4)/(x^2-4) = oo#

Note it takes a lot longer to explain than it does to do!