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

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

#### Explanation:

As $x \rightarrow 2$, the numerator goes to $8$ (which is not $0$) and the denominator goes to $0$.

This form $\frac{\text{non-} 0}{0}$ tells us that the function is increasing or decreasing without bound on each side. (The one-sided limits are $\pm \infty$)

To see which is happening as $x \rightarrow {2}^{+}$, 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.

${\lim}_{x \rightarrow {2}^{+}} \frac{{x}^{2} + 4}{{x}^{2} - 4} = \infty$

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