Question

Evaluate the given limits of the piecewise defined functions \(f\). $$ \begin{array}{ll} f(x)=\left\\{\begin{array}{cc} 2 x^{2}+5 x-1 & x<0 \\ \sin x & & x \geq 0 \end{array}\right. \\ \begin{array}{ll} \text { (a) } \lim _{x \rightarrow 0^{-}} f(x) & \text { (c) } \lim _{x \rightarrow 0} f(x) \\ \text { (b) } \lim _{x \rightarrow 0^{+}} f(x) & \text { (d) } f(0) \end{array} \end{array} $$

Short answer

(a) -1, (b) 0, (c) does not exist, (d) 0.

Step-by-step solution

Evaluate Left-Hand Limit

To find \( \lim_{x \to 0^{-}} f(x) \), we need to consider the function definition for values of \( x \) less than 0. Thus, \( f(x) = 2x^2 + 5x - 1 \). Substitute \( x \to 0 \) in this expression to evaluate the limit: \ \( \lim_{x \to 0^{-}} f(x) = 2(0)^2 + 5(0) - 1 = -1 \).

Evaluate Right-Hand Limit

Next, we find \( \lim_{x \to 0^{+}} f(x) \). For values of \( x \) greater than or equal to 0, \( f(x) = \sin x \). Substitute \( x \to 0 \) to evaluate this limit: \( \lim_{x \to 0^{+}} f(x) = \sin(0) = 0 \).

Evaluate Two-Sided Limit

For \( \lim_{x \to 0} f(x) \), we need both the left-hand limit \( \lim_{x \to 0^{-}} f(x) \) and the right-hand limit \( \lim_{x \to 0^{+}} f(x) \) to be equal. From Steps 1 and 2, we have \(-1\) and \(0\) respectively. Since these are not equal, \( \lim_{x \to 0} f(x) \) does not exist.

Evaluate \( f(0) \)

Finally, evaluate \( f(0) \) directly from the piecewise function definition. Since \( x = 0 \) falls in the second function's domain, use \( f(x) = \sin x \). Thus, \( f(0) = \sin(0) = 0 \).

Key concepts

Left-Hand Limit

When dealing with the left-hand limit, we're interested in what happens to the function as we approach a particular point from the left.
In mathematical terms, if we say we're evaluating the left-hand limit of a function at a point, we mean \lim_{x \to a^{-}} f(x)\.
Here, we specifically looked at the piecewise function as \(x\) approaches zero from the negative side.For the function \(f(x)\):

• If \(x < 0\), we use \(2x^2 + 5x - 1\)
• As \(x\) gets closer to zero, we evaluate \( \lim_{x \to 0^{-}} (2x^2 + 5x - 1) \)

Plugging \(x = 0\) into \(2x^2 + 5x - 1\) gives \(-1\). So, the left-hand limit is \(-1\).
This step involves substituting zero into the polynomial, while keeping within the specified range of \(x < 0\).
Understanding this approach helps us determine the behavior of piecewise functions from one side.

Right-Hand Limit

To assess the right-hand limit, we examine the behavior of a function as it approaches a specific point from the right. For this, we learn about \( \lim_{x \to a^{+}} f(x) \).
This concept is crucial as it provides insights about the function's behavior beyond the given point.In our piecewise example:

• If \(x \geq 0\), the function is \( \sin x \)
• Approaching zero from the right involves evaluating \( \lim_{x \to 0^{+}} \sin x \)

Since \( \sin(0) = 0 \), the right-hand limit is \(0\).
It's important to note that the right-hand limit selects the value corresponding to the function's definition at or past the boundary point.
This method ensures we've captured the necessary behavioral aspect of the piecewise function accurately from the right side.

Two-Sided Limit

A two-sided limit combines both the left-hand and right-hand limits at a specific point. We value the expression \( \lim_{x \to a} f(x) \), where it's vital for both lateral limits to coincide.
This process confirms continuity or identifies discontinuities in a function.Let's revisit our function:

• The left-hand limit yields \(-1\)
• The right-hand limit provides \(0\)

Since \(-1\) \(eq\) \(0\), the two-sided limit \( \lim_{x \to 0} f(x) \) does not exist.
Two-sided limits serve as a valuable tool for checking whether a function behaves consistently from both ends at any given point.
If they do not match, like in our piecewise function, we identify a break or jump, indicating a discontinuity.

Piecewise Function Evaluation

Evaluating a piecewise function involves assessing its behavior at designated intervals within its domain.
Such a function is defined by different expressions depending on the value of \(x\). Our example illustrates this with:

• \(2x^2 + 5x - 1\) for \(x < 0\)
• \(\sin x\) for \(x \geq 0\)

To evaluate \(f(0)\), we use the definition applicable at \(x = 0\), which in this case, is \(\sin x\).
Thus, \(f(0) = \sin(0) = 0\).
This example shows that piecewise functions may pivot at key points, holding unique expressions occasionally resulting in non-existing two-sided limits.
Mastering piecewise evaluation facilitates a greater understanding of function behaviors at any given section of their domains.