Question

In Exercises \(55 - 58 ,\) complete parts \(( a ) - ( d )\) for the piecewise- definedfunction. \(\quad (\) a) Draw the graph of \(f\) . (b) At what points \(c\) in the domain of \(f\) does \(\lim _ { x \rightarrow c } f ( x )\) exist? (c) At what points \(c\) does only the left-hand limit exist? (d) At what points \(c\) does only the right-hand limit exist? $$f ( x ) = \left\\{ \begin{array} { l l } { \sqrt { 1 - x ^ { 2 } } , } & { 0 \leq x < 1 } \\ { 1 , } & { 1 \leq x < 2 } \\ { 2 , } & { x = 2 } \end{array} \right.$$

Short answer

The graph consists of a half-circle on the interval \( [0,1)\), a horizontal line at \(y=1\) on the interval \([1,2)\), and a point at \(x=2\), \(y=2\). The limit exists at every point except potentially \(x=1\) and \(x=2\). The left-hand limit exists and equals 0 at \(x=1\), and doesn't exist at \(x=2\). The right-hand limit exists and equals 1 at \(x=1\), and doesn't exist at \(x=2\).

Step-by-step solution

Graph the Function

It's necessary to plot three separate pieces of the function on the graph: the half-circle function \(f(x) = \sqrt { 1 - x ^ { 2 } }\) for \( 0 \leq x < 1\), the horizontal line \(f(x) = 1 \) for \( 1 \leq x < 2\), and the point at \(x = 2\) where \(f(x) = 2\). The three pieces should be combined into one graph.

Identify Where Limits Exist

The limits exist at every point \(c\) where \(f(x)\) is defined, except potentially where \(f(x)\) jumps from one piece to another (at \(x=1\) and \(x=2\)). The limit at these points need to be evaluated by taking the right-hand and left-hand limit.

Evaluate the Left-Hand Limit

A left-hand limit only exists at \(x = 1\) and \(x = 2\), as these are the points where \(f(x)\) jumps from one piece to another. For \(x = 1\), the left-hand limit is the limit of \(f(x) = \sqrt { 1 - x ^ { 2 } }\) as \(x\) approaches 1 from the left, which is \( \sqrt{1-1} = 0\). For x = 2, there's no left-hand limit since there's no piece of the function defined for x > 2.

Evaluate the Right-Hand Limit

A right-hand limit only exists at \(x = 1\). This is determined by the limit of \(f(x) = 1 \) as \(x\) approaches 1 from the right, which is 1. At \(x = 2\), there's no right-hand limit since that's the extreme right point where the function is defined.