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!