Consider the following graph of the function f. The x y coordinate plane is given. The function labeled y = f(x) with 5 parts is graphed. The first part is a curve beginning at the open point (−3, 2), goes down and right becoming less steep, changes directions at the approximate point (−1, 1), goes up and right becoming more steep, sharply changes direction at (0, 2), goes horizontally right, and ends at the open point (2, 2). The second part is the closed point (2, 1). The third part is a curve beginning at the open point (2, 2), goes up and right becoming less steep, and ends at the vertical asymptote x = 4. The fourth part is the closed point (4, 2). The fifth part is a curve entering the window at the top of the first quadrant to the right of the vertical asymptote x = 4, goes down and right becoming less steep, and exits the window in the first quadrant just above the x−axis. Find the following one-sided limits. (If an answer does not exist, enter DNE.) lim x→0− f(x) = lim x→0+ f(x) = Determine whether the statement is true or false. lim x→0 f(x) = 1 True False

According to the description of the function y = f(x) from the point (-1,1) it goes up and right becoming more steep, sharply changes direction at (0,2), goes horizontally right, and ends at the open point (2,2).

In this part we can assume that the point (0,2) belongs to f(x). Actually, we can say that when x=0, f(x)=2. And this is important because of the limit of f(x) when it approaches to 0 from the left (0-) or the right (0+).

Since the point (0,2) is the end of the road from the point (-1,1) we can say the limit when x approaches to zero, from the left, is 2.

[tex]\lim_{x \to 0^{-} } f(x)=2[/tex]

And since the same point (0,2) is the beginning of the road for when it goes horizontally right to the (2,2) point we can say that the limit when x approaches to zero, from the right, is 2.

[tex]\lim_{x \to 0^{+} } f(x)=2[/tex]

So the limit when x approaches to zero is 2, not 1 therefore the statement is false.

And we can write:

[tex]\lim_{x \to 0^{-} } f(x)=\lim_{x \to 0^{+} } f(x)=2[/tex]