# Determining Limits Graphically

## Key Questions

• I am not sure if there is a TI-84 Plus function that directly finds the value of a limit; however, there is a way to approximate it by using a table. Let us approximate the value of the limit

${\lim}_{x \to 1} \frac{\sqrt{x + 3} - 2}{x - 1}$

Step 1: Go to "Y=", then type in the function. Step 2: Go to "TBL SET" (2nd+WINDOW), then set TblStart=.97 and $\Delta$Tbl=.01. (Note: TblStart is the starting x-value in the table, so put a number slightly smaller that the number x approaches. $\Delta$Tbl is the increment value in the x-column, so make it sufficiently small for the precision you need.)

Step 3: Go to "" TABLE (2nd+GRAPH). As you can see in the table above, the function value (${Y}_{1}$) approaches 0.25 (or 1/4) as x approaches 1; therefore, we conclude that

${\lim}_{x \to 1} \frac{\sqrt{x + 3} - 2}{x - 1} = \frac{1}{4}$

• If you're using a graph to find this limit, the first thing you'll want to do is graph the function.

$f \left(x\right) = {x}^{2} + 2$ is a parabola that looks like this: If you want to find out how to graph this, you can either draw the graph of a normal parabola and translate it vertically by two units upwards (2 is being added to the ${x}^{2}$, which is why it goes up), or you can create a table of values and plug in input $x$ values to get output $y$ and you'll get an idea of the shape of the graph.

Now we're interested in knowing what is happening at $x$=5. Luckily, the function is defined there. If we look at the graph, at $x$=5, y=27. It's a little bit hard to tell on the graph because of the exponentially increasing y-values, but we know that $y$=27 because $y = \left({5}^{2} + 2\right) = 27$. We can plug in $x$ directly to find the limit because the function is defined and continuous there.

To get an idea of an it intuitively means to find a limit on a graph though, you can look at the graph and decide what is happening at $x$=5. When you move to $x$=5 from the right, what is $y$ tending to? Well, to 27. Also, when you move to $x$=5 from the left, what is $y$ tending it? 27 again. You can think of limits from the right and left as arrows pointing right and left respectively to the $x$ value you're looking for. You're kind of trying to "pin point" what is exactly is happening at your graph at that given $x$ point. In this case it's quite simply reading the $y$ value off the graph, since the left and right limits tend to the same point and therefore are equal.

So,

${\lim}_{x \to 5} \left({x}^{2} + 2\right) = 27$