Estimação de limites por uma boa aproximação numérica

First, let's explore the limit of a nice, continuous function, $f(x) = \frac{1}{2}x$. If we try to find the limit of a function at a point for which the function is defined, then the limit at that point is equivalent to the function evaluated at that point.

For our example using $f(x) = \frac{1}{2}x$, this means that $\lim_{x\to2} \frac{1}{2}x = f(2) = \frac{1}{2} \times 2 = 1$.

Intuitively, the limit describes the behavior of a function as we approach a certain point from both sides. If the function is defined at the point we choose, then the function approaches the limit and happens to equal it at the point at which the limit is taken.

There are also cases in which the point at which you are taking a limit also happens to be an endpoint of the domain or the endpoint of a discontinuity. As long as the point is included in the domain, then it is not an issue: $$\lim_{x\to0} \sqrt{x} = 0$$

But, suppose that the point we're looking for isn't defined by our function. What if the point at which we are looking for the limit lies outside of the domain?

From the plot, it is still clear that $\ln{0}$ is not defined. However, $\ln{0.1}$ is defined. Conveniently, so are $\ln{0.01}$, $\ln{0.001}$, $\ln{0.0001}$, $\ln{0.00001}$, and so on (although that might not be apparent from the plot). These five natural logarithms approximate to $-2.30259$, $-4.60517$, $-6.90776$, $-9.21034$, and $-11.5129$, respectively. Eyeballing these results, its clear that as we approach $x = 0$, $y = \ln{x}$ is strictly decreasing. Furthermore, the rate at which the function decreases seems to be increasing. The sharp drop at the left end of the function in the plot supports that conclusion.

If we plot sufficiently enough points, we'll find that $\ln{x}$ decreases without bound as $x$ approaches $0$ (if you're not convinced, plug in some super tiny numbers into $\ln$ on a calculator). A more succinct way of expressing this would be: $$\lim_{x\to0} (\ln{x}) = -\infty$$

At first, it may not make immediate sense why the answer is $5$. After all, the rational function $\frac{x^2 + x - 6}{x - 2}$ isn't even defined at $x = 2$. How can the limit exist?

Plotting the function once again proves to be helpful in understanding how limits work:

The function conveniently simplifies down to $x + 3$ across it's entire domain, except for when $x = 2$. However, the function still simplifies down to $x + 3$ when we get extremely close to $x = 2$. Like we did with the previous example with $ln$, we can approximate the values of the function as we approach $x = 2$.

Because $x = 2$ is not an "endpoint" of the domain, it is important that we approach $x = 2$ from both sides to make sure that the limits on both sides are consistent. If we were to end up with different values when we approach $x = 2$ from either side, then the limit at $x = 2$ would not exist because both the left and right side limits must be the same in order for the limit to be defined.

As we approach $x = 2$ from the left and right sides, it becomes clear that we are tracing the line $y = x + 3$. Even though the function is not defined at $x = 2$, we can then say that $$\lim_{x\to2} \frac{x^2 + x - 6}{x - 2} = 2 + 3 = 5$$ because the function gets infinitely close to $5$ as we get infinitely close to $x = 2$.

Next, let's look at $\lim_{x\to0} |\frac{1}{x}|$. The plot of $\frac{1}{x}$ looks something like

Intuitively, the limit would be $+\infty$. After all, the graph seems to shoot off towards $+\infty$ as we approach from both the left and right sides. $+\infty$ is indeed the answer that most people would agree upon.

However, some math teachers and mathematicians argue that a two-sided limit to $\pm \infty$ is incorrect. The argument is that a function is not guaranteed to increase or decrease without bound at the same rate on the left and right sides of the limit. If we were to follow this premise, then $\lim_{x\to0} |\frac{1}{x}|$ does not exist. Because $|\frac{1}{x}|$ is symmetrical about the $x$-axis, our answer of $+\infty$ is somewhat more justifiable, but the use of $\pm \infty$ as an answer in two-sided limits will nevertheless generate some dialogue in some cases, especially when the rate of growth/decay on either side of the limit to $\pm \infty$ differs.

If we plot the function $\frac{1}{x}$, it becomes immediately clear why the limit at $0$ does not exist.

We can see pretty easily that $\lim_{x\to0^+} \frac{1}{x} = +\infty$ and $\lim_{x\to0^-} \frac{1}{x} = -\infty$. However, because $\lim_{x\to0^+} \frac{1}{x} \ne \lim_{x\to0^-} \frac{1}{x}$, $\lim_{x\to0} \frac{1}{x}$ does not exist.

From the graph, we can see that our conclusion makes sense. To say that the function definitively goes to either $+\infty$ or $-\infty$ would be incorrect. Claiming the limit to be $\pm\infty$ would also be incorrect, because the limit can only take on one answer.

This is also a handy explanation of why we cannot divide by $0$: as we divide by progressively smaller positive numbers (approaching $0$), our quotient increases without bound to $+\infty$, yet as we divide by progressively larger negative numbers (approaching $0$), our quotient instead decreases without bound to $-\infty$.

(Video) Estimate a Limit Numerically

by YourCalculusHelper

Here's a video by YourCalculusHelper showing you how you estimate a limit numerically.

If you can't tell where the limit is heading you can always find the average of the two values to the left and right of the number and then add it onto the smaller of the two.

For example, in the video $f(3.999)=0.20004$ and $f(4.001)=0.19996$. So take the average of the two, $\frac{0.20004-0.19996}{2}=0.00004$. Now either add this value onto the smaller of the two (in this example it would be $0.19996$) or subtract the value from the larger of the two ($0.20004$), and you get $0.19996+0.00004=0.2$