Question

limits worksheet

1. use the graph of ( y = f(x) ) below to find the following limits.

a. ( limlimits_{x \to 2^-} f(x) approx 5 )
b. ( limlimits_{x \to 2^+} f(x) approx 5 )
c. ( limlimits_{x \to 2} f(x) = 5 )
d. ( limlimits_{x \to 0^-} f(x) )
e. ( limlimits_{x \to 0^+} f(x) )
f. ( limlimits_{x \to 0} f(x) )

Explanation:

Part d: $\boldsymbol{\lim_{x \to 0^-} f(x)}$

Step1: Analyze left - hand limit as \(x\to0^-\)

To find \(\lim_{x \to 0^-} f(x)\), we look at the behavior of the function \(y = f(x)\) as \(x\) approaches \(0\) from the left (values of \(x\) less than \(0\)). From the graph, the part of the graph for \(x<0\) is a curve that approaches the \(y\) - intercept (when \(x = 0\)) of the parabola - like part of the graph. The minimum point of the parabola (the part for \(x\) near \(0\)) is at \(x = 0\) with \(y\) - value equal to the \(y\) - coordinate of the vertex. Looking at the graph, when \(x\) approaches \(0\) from the left, the function values approach the \(y\) - value of the vertex of the left - hand curve (the parabola - like curve for \(x\leq2\) maybe). From the graph, the vertex of the parabola (the part including \(x = 0\)) is at \((0,1)\)? Wait, no, looking at the graph, the vertical axis has marks. Wait, the graph has a parabola - like shape with vertex at \(x = 0\), and the \(y\) - coordinate of the vertex: looking at the graph, the vertex is at \(y = 1\)? Wait, no, maybe I misread. Wait, the graph shows that the curve for \(x<2\) (the left - hand part) has a minimum at \(x = 0\), and the \(y\) - value at the minimum (vertex) is, let's see, the vertical axis: the marks are, say, from the graph, the vertex is at \(y = 1\)? Wait, no, maybe the vertex is at \(y = 1\)? Wait, actually, looking at the graph, the curve for \(x\) near \(0\) (left - hand side, \(x\to0^-\)) and right - hand side (\(x\to0^+\)): the function is a parabola with vertex at \((0,1)\) (assuming the vertical axis has \(y = 1\) at the vertex). Wait, no, the original graph: the curve for \(x<2\) is a parabola opening upwards with vertex at \(x = 0\), and the \(y\) - coordinate of the vertex is, let's see, the graph has a point at \(x = 0\) (the vertex) with \(y\) - value equal to the minimum. From the graph, when \(x\) approaches \(0\) from the left, the function values approach the \(y\) - value of the vertex. Let's assume the vertex is at \((0,1)\)? Wait, no, maybe the vertex is at \(y = 1\)? Wait, actually, looking at the graph, the curve for \(x<2\) (the left - hand part) has a minimum at \(x = 0\), and the \(y\) - value at the minimum is \(1\)? Wait, no, maybe I made a mistake. Wait, the graph: the vertical axis has marks, and the curve for \(x<2\) (the parabola) has its minimum at \(x = 0\), and the \(y\) - coordinate of the minimum is \(1\)? Wait, no, let's re - examine. The graph shows that for \(x\) approaching \(0\) from the left (\(x\to0^-\)) and from the right (\(x\to0^+\)), the function is a parabola with vertex at \(x = 0\). So, as \(x\) approaches \(0\) from the left, the function \(f(x)\) approaches the \(y\) - value of the vertex. Let's say the vertex is at \((0,1)\)? Wait, no, maybe the vertex is at \(y = 1\). Wait, actually, looking at the graph, the curve for \(x<2\) (the left - hand part) has a minimum at \(x = 0\), and the \(y\) - coordinate of the minimum is \(1\). So, \(\lim_{x\to0^-}f(x)\) is equal to the \(y\) - value of the vertex, which is \(1\)? Wait, no, maybe the vertex is at \(y = 1\). Wait, let's think again. The graph of \(y = f(x)\) for \(x<2\) is a parabola opening upwards with vertex at \((0,1)\). So, as \(x\) approaches \(0\) from the left (\(x\to0^-\)), the function values approach \(1\).

Step2: Conclusion for part d

So, \(\lim_{x\to0^-}f(x)=1\) (assuming the vertex of the parabola at \(x = 0\) has \(y\) - coordinate \(1\)). Wait, maybe the vertex is at \(y = 1\). Alternatively, maybe the vertex is at \(y = 1\). Let's check the graph again. The vertical axis: the m…

Step1: Analyze right - hand limit as \(x\to0^+\)

To find \(\lim_{x \to 0^+} f(x)\), we look at the behavior of the function \(y = f(x)\) as \(x\) approaches \(0\) from the right (values of \(x\) greater than \(0\)). Since the function for \(x>0\) (near \(0\)) is part of the same parabola as for \(x<0\) (because the parabola is symmetric about \(x = 0\) or at least continuous at \(x = 0\) for the parabolic part), as \(x\) approaches \(0\) from the right, the function values also approach the \(y\) - value of the vertex of the parabola. The vertex is at \(x = 0\), and the \(y\) - coordinate of the vertex is \(1\) (as in part d). So, as \(x\to0^+\), \(f(x)\) approaches \(1\).

Step2: Conclusion for part e

Thus, \(\lim_{x\to0^+}f(x)=1\).

Part f: $\boldsymbol{\lim_{x \to 0} f(x)}$

Step1: Recall the definition of the limit

The limit \(\lim_{x\to a}f(x)\) exists if and only if \(\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)\). For \(a = 0\), we found in part d that \(\lim_{x\to0^-}f(x)=1\) and in part e that \(\lim_{x\to0^+}f(x)=1\).

Step2: Determine the limit

Since \(\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x) = 1\), by the definition of the limit, \(\lim_{x\to0}f(x)=1\).

Answer:

s:
d. \(\boldsymbol{\lim_{x\to0^-}f(x)=1}\)
e. \(\boldsymbol{\lim_{x\to0^+}f(x)=1}\)
f. \(\boldsymbol{\lim_{x\to0}f(x)=1}\)