Determining Limits Graphically

Key Questions

• How do you find limits on a graphing calculator?

  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}{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}{sqrt{x+3}-2}/{x-1}=1/4`

  Wataru · 4 · Sep 24 2014
• How do you find `lim_(x->5)(x^2+2)` using a graph?

  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`

  Sally · · Sep 27 2014

• How do you find `lim_(x->5)(x^2+2)` using a graph?
• How do i graph limits?
• How do you find limits on a graphing calculator?
• How do you use a graph to determine limits?
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