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

Sep 27, 2014

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,

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