Question

Find the limit \(L\). Then use the \(\varepsilon-\delta\) definition to prove that the limit is \(L\). $$ \lim _{x \rightarrow-4}\left(\frac{1}{2} x-1\right) $$

Short answer

The limit \(L\) of the expression as \(x \rightarrow -4\) is -3. It can be verified using the \(\varepsilon -\delta\) definition by choosing \(\delta = 2 \varepsilon\).

Step-by-step solution

Compute the limit

To find the limit of the expression \( \frac{1}{2} x-1 \) as \( x \rightarrow -4 \), just replace \( x \) with \( -4 \) in the expression: \(L = \frac{1}{2}*(-4) - 1 = -2 - 1 = -3 \)

Understand the \(\varepsilon -\delta \) definition

The \(\varepsilon -\delta \) definition states that the limit of function \(f(x)\) approaches \(L\) as \(x\) approaches \(c\) if for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that whenever \(0 < |x-c| < \delta \), it follows that \( |f(x)-L| < \varepsilon \).

Formulate in terms of the given function

In this case, \(f(x)=\frac{1}{2} x-1\), \(c=-4\), and \(L=-3\). Therefore, for a given \(\varepsilon > 0\), we want to find a \(\delta > 0\) such that if \(0 < |x-(-4)| = |x+4| < \delta\), then \(|\frac{1}{2}x-1 - (-3)| = |\frac{1}{2}x+2| < \varepsilon \).

Solve for \(\delta \)

To ensure that \( |\frac{1}{2}x+2| < \varepsilon \), we can choose \(\delta = 2 \varepsilon \). This way, if \(|x+4| < \delta\), we have \(|\frac{1}{2}x+2| = \frac{1}{2}|x+4| < \varepsilon \).

Key concepts

Limit of a Function

Understanding the limit of a function is crucial in calculus. It describes the behavior of a function as the input approaches a certain value, without necessarily attaining that value. For instance, when we say \( \lim _{x \rightarrow c}f(x) = L \), we mean that as \(x\) gets closer to \(c\), the value of \(f(x)\) gets arbitrarily close to \(L\).

In the exercise provided, we find the limit of the function \( \frac{1}{2} x-1 \) as \( x \) approaches -4. The process involves simply substituting the -4 in place of \(x\) and calculating the resulting value which equals -3. This straightforward substitution method works well when the function is well-behaved at the point we're interested in. However, not all limits are this straightforward—some may involve more complex functions or situations where direct substitution is not possible. That's where other techniques, like the epsilon-delta definition, come into play.

Continuous Function

A continuous function is a function that has no interruptions, jumps, or breaks in its graph. Formally, a function \(f\) is said to be continuous at a point \(c\) if \( \lim _{x \rightarrow c}f(x) = f(c) \). This means the limit of \(f\) as \(x\) approaches \(c\) from any direction equals the value that \(f\) actually takes at \(c\).

Using the epsilon-delta definition, one can rigorously prove that a function is continuous at a certain point. For the solved exercise, proving continuity at \(x = -4\) involves showing that for every \( \varepsilon > 0 \), there is a \( \delta > 0 \) such that whenever \(0 < |x - (-4)| < \delta \) it follows that \( |\frac{1}{2}x - 1 - (-3)| < \varepsilon \). Since we were able to define a \( \delta \) that meets this condition, it indicates that the function \( \frac{1}{2} x-1 \) is continuous at \(x = -4\).

Limit Laws

Limit laws are a set of rules that allow us to compute the limits of functions in a piecewise manner, which is often more straightforward than trying to handle the entire function at once. These laws include the sum law, product law, quotient law, and others. They state that the limit of a sum is the sum of the limits, the limit of a product is the product of the limits, and so on, provided that the limits exist.

Using limit laws, one can break down complex functions into simpler parts to evaluate their limits. For the function in our exercise, we could also use the limit law for the subtraction of functions if it were more complex. However, since direct substitution was feasible, the need for this process wasn’t necessary. The epsilon-delta proof, in turn, affirms our result, ensuring that our limit calculation stands on a solid mathematical foundation.