Problem 44 - Precalculus: Mathematics for Calculus

Question

Does the Limit Exist? Let $$h(x)=\left\\{\begin{array}{ll} x & \text { if } x<0 \\ x^{2} & \text { if } 02 \end{array}\right.$$ (a) Evaluate each limit if it exists. (i) \(\lim _{x \rightarrow 0^{+}} h(x)\) (iv) \(\lim _{x \rightarrow 2^{-}} h(x)\) (ii) \(\lim _{x \rightarrow 0} h(x)\) (v) \(\lim _{x \rightarrow 2^{+}} h(x)\) (iii) \(\lim _{x \rightarrow 1} h(x)\) (vi) \(\lim _{x \rightarrow 2} h(x)\) (b) Sketch the graph of \(h.\)

Step-by-Step Solution

Answer

Limits: (i) 0, (ii) 0, (iii) 1, (iv) 4, (v) 6, (vi) does not exist.

Step 1: Analyze the function for x -> 0+

For the limit as \(x\) approaches 0 from the right \((0^+)\), we examine the piece of the piecewise function where \(0 < x \leq 2\). In this interval, \(h(x) = x^2\). Thus, \[\lim_{x \to 0^+} h(x) = \lim_{x \to 0^+} x^2 = 0^2 = 0.\]

Step 2: Analyze the function for x -> 0-

For the limit as \(x\) approaches 0 from the left \((0^-)\), we use the piece where \(x < 0\), where \(h(x) = x\). Therefore:\[\lim_{x \to 0^-} h(x) = \lim_{x \to 0^-} x = 0.\]

Step 3: Conclude the limit as x -> 0

Since \[\lim_{x \to 0^+} h(x) = \lim_{x \to 0^-} h(x) = 0,\]the two one-sided limits are equal; thus:\[\lim_{x \to 0} h(x) = 0.\]

Step 4: Analyze the function for x -> 1

For \(x \to 1\), we only consider the interval where \(0 < x \leq 2\), as 1 falls within this range. Thus, \(h(x) = x^2\). We calculate:\[\lim_{x \to 1} h(x) = \lim_{x \to 1} x^2 = 1^2 = 1.\]

Step 5: Analyze the function for x -> 2-

For the limit as \(x\) approaches 2 from the left \((2^-)\), \(x\) falls in the interval \(0 < x \leq 2\), and \(h(x) = x^2\). Thus:\[\lim_{x \to 2^-} h(x) = \lim_{x \to 2^-} x^2 = 2^2 = 4.\]

Step 6: Analyze the function for x -> 2+

For the limit as \(x\) approaches 2 from the right \((2^+)\), \(x > 2\), which implies \(h(x) = 8 - x\). Thus:\[\lim_{x \to 2^+} h(x) = \lim_{x \to 2^+} (8 - x) = 8 - 2 = 6.\]

Step 7: Conclude the limit as x -> 2

Since \[\lim_{x \to 2^-} h(x) = 4\] and \[\lim_{x \to 2^+} h(x) = 6\] are not equal, the limit as \(x\) approaches 2 does not exist.

Step 8: Sketch the graph of h(x)

Plot the function according to each interval:1. For \(x < 0\), \(h(x) = x\), which is a line through the origin with slope 1.2. For \(0 < x \leq 2\), \(h(x) = x^2\), a parabolic shape starting from 0 at \(x=0\) and reaching 4 at \(x=2\).3. For \(x > 2\), \(h(x) = 8-x\), a decreasing line starting from 6 at \(x=2\).

Key Concepts

Piecewise Functions

A piecewise function is one that has different expressions or formulas for different parts of its domain. Seems complex, right? But it gets easier once you grasp the concept. Just think of it as a function wearing different hats for different intervals of values for the variable \(x\). In our given function \(h(x)\), it has three different rules:

• For \(x < 0\), \(h(x) = x\). So, whenever \(x\) is less than 0, we just use the rule \(h(x) = x\).
• For \(0 < x \leq 2\), \(h(x) = x^2\). This applies when \(x\) is between 0 and 2, including 2.
• For \(x > 2\), \(h(x) = 8 - x\). In this scenario, once \(x\) is greater than 2, we switch to using \(8 - x\).

Paying attention to these intervals is key. That’s how you evaluate limits or solve equations involving piecewise functions. It's all about finding the right part of the function to "put on". Once you're comfortable, these types of functions become manageable.

Continuity

In mathematics, continuity is a property that describes whether a function has any interruptions or jumps at any given point in its domain. Imagine you’re drawing a graph without lifting your pencil; that's a nice visual of a continuous function.

In examining the function \(h(x)\), you’ll see that:

• The limit as \(x\) approaches 0 (from both directions) exists and equals \(0\). Hence, \(h(x)\) is continuous at \(x = 0\).
• For \(x = 2\), the left-hand limit is \(4\) and the right-hand limit is \(6\). Both don't match, signaling a break or jump in continuity here.

So, while \(h(x)\) is continuous at \(x = 0\), it's not at \(x = 2\) because of the different behaviors from the left and right sides. Understanding continuity helps in predicting how functions behave and whether they’re stable or have disruptions.

Graph Sketching

Sketching graphs can turn abstract numbers into visual stories. It tells us so much about a function instantly: areas of increase or decrease, peaks, valleys, and points of discontinuity are all out in the open.

Here's how to tackle the graph of \(h(x)\):

• For \(x < 0\), plot a line that passes through the origin with a slope of 1 (because \(h(x) = x\)).
• From \(0 < x \leq 2\), sketch out a parabola that begins at \(h(x) = 0\) when \(x = 0\) and reaches up to \(h(x) = 4\) when \(x = 2\).
• Finally, for \(x > 2\), draw a line starting at \(h(x) = 6\) when \(x = 2\), which decreases linearly (since \(h(x) = 8 - x\)).

These segments, though distinct in the graph, represent the simple changes in direction or steepness dictated by each part of our piecewise function. Drawing it lets us see the jumps and the smooth transitions, making the function more tangible.