Question

Find the limit \(L\). Then use the \(\varepsilon-\delta\) definition to prove that the limit is \(L\). $$ \lim _{x \rightarrow-3}\left(x^{2}+3 x\right) $$

Short answer

The limit \(L\) of \(x^{2}+3x\) as \(x\) approaches -3 is 0. This limit is proved using the \(\varepsilon-\delta\) definition by showing that for every \(\varepsilon > 0\), we can find a \(\delta > 0\) such that for every \(x\) with \( 0 < |x - (-3)| < \delta \), the value of \( |f(x) - L| < \varepsilon \).

Step-by-step solution

Find the Limit

The limit is found by substituting the limiting value into the function, which gives: \[ \lim _{x \rightarrow-3}\left(x^{2}+3 x\right) = (-3)^{2}+3(-3) = 9-9 = 0 \] So, \(L = 0\).

Define the Function and Limit

After finding the limit, the next step is to define \(f(x)\) and \(L\) in terms of \(x\), \(\delta\), and \(\varepsilon\). These are: \(f(x) = x^{2}+3x\) and \(L = 0\). The goal is to prove that for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that if \( 0 < |x - (-3)| < \delta \), then \( |f(x) - L| < \varepsilon \).

Use the Epsilon-Delta Definition

We need to express \(\delta\) in terms of \(\varepsilon\), such that the condition \( |f(x) - L| < \varepsilon \) is satisfied whenever \( 0 < |x - (-3)| < \delta \) is true. Start with \( |f(x) - L| < \varepsilon \), which simplifies to \( |x^{2}+3x| < \varepsilon \). Rewrite this as \( |x||x+3| < \varepsilon \) (because the factorization of \(x^2 + 3x = x(x+3)\)), and note that \( |x+3| = |x - (-3)| < \delta \). For \(x\) close to -3, \( |x| \) is also going to be close to 3, so using the number 3 to get an upper estimation is a good choice - so say, \( |x| \leq 3 \). Thus, select \( \delta = \min \left( 1, \frac{\varepsilon}{3} \right) \). So whenever \( 0 < |x - (-3)| < \delta \), we'll have \( |f(x) - L| = |x||x+3| \leq 3 \delta \leq \varepsilon \), thus verifying the limit.

Key concepts

Limit of a Function

In calculus, the concept of a limit is pivotal for analyzing how functions behave as they approach a certain point, without necessarily reaching it. For instance, the exercise we have above asks us to find the limit of the function as x approaches -3. This is formally written as \(\lim _{x \rightarrow -3}\left(x^{2}+3 x\right)\).

By directly substituting the value of -3 into the function, we find that the limit L equals 0. This substitution can work when the function is well-behaved (i.e., continuous) at the point in question. However, to rigorously prove that this is indeed the correct limit, we employ the \(\varepsilon-\delta\) definition of a limit. This formal approach ensures that for every infinitesimally small positive distance (\(\varepsilon\)), we can find a corresponding range (\(\delta\)) around -3 such that the function's value is guaranteed to be within \(\varepsilon\) of 0 whenever x is within \(\delta\) of -3.

Continuity

Continuity is a quality of functions that allows them to be predictable and well-behaved within certain intervals or even across their entire domain. A function f(x) is said to be continuous at a point a if \(\lim _{x \rightarrow a}f(x) = f(a)\). In other words, you can draw the function without lifting your pen off the paper. The concept of continuity closely ties into limits because a limit must exist at a point for the function to be continuous there.

Our example function, f(x) = x^2 + 3x, is a polynomial, and polynomials are continuous everywhere on their domain. This means that to prove the limit, it's sufficient to show that for any arbitrarily small \(\varepsilon\) (representing the vertical distance), we can find a \(\delta\) (representing the horizontal distance) such that the values of f(x) are as close to L as we want when x is near the point of interest—in this case, -3.

Calculus

Calculus is a branch of mathematics dealing with rates of change and the accumulation of quantities. The two main types are differential calculus, concerning instantaneous rates of change, and integral calculus, focusing on the accumulation of quantities. This study of limits is a fundamental concept in differential calculus, as it leads to the definition of the derivative, a core tool used for understanding how functions change.

In our \(\varepsilon-\delta\) proof, we are using the bounds set by \(\varepsilon\) to define an interval around a point where the function's behavior can be predicted with precision. Doing this is core to calculus, as it allows for the abstraction of complex, real-world problems into models that can be analysed and solved. The ability to prove limits exists confirms the reliability of these mathematical models. By mastering these skills, one can better understand the natural phenomena that calculus helps to describe.