Question

How do you find `lim_(x->5)(x^2+2)` using a graph?

Answer 1

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

`f(x)=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=(5^2+2)=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->5) (x^2+2)=27`