Problem 26 - Calculus Early Transcendentals

Question

Suppose \(\lim _{x \rightarrow a} f(x)=L\). Prove that \(\lim _{x \rightarrow a}[c f(x)]=c L,\) where \(c\) is a constant.

Step-by-Step Solution

Answer

Question: Prove that the limit of a function multiplied by a constant is equal to the limit of the function itself multiplied by the same constant, i.e., \(\lim_{x\rightarrow a} [c f(x)] = c (\lim_{x\rightarrow a} f(x))\), where \(f(x)\) is a function and \(c\) is a constant. Answer: Following the step by step solution given in the explanation, we've successfully shown that if the limit of \(f(x)\) as \(x\) approaches \(a\) is \(L\), then the limit of \(cf(x)\) as \(x\) approaches \(a\) is equal to \(cL\). By proving this assertion using the definition of limit and appropriate algebraic manipulations, we have demonstrated the required result: \(\lim_{x\rightarrow a} [c f(x)] = c (\lim_{x\rightarrow a} f(x))\).

Step 1: Write the function as a product of the constant and the original function

Given a function, \(f(x)\), we are asked to find the limit of the product \(cf(x)\), where \(c\) is a constant. This can be written as: \(\lim_{x\rightarrow a} [c f(x)]\)

Step 2: Write the definition of the limit for the given function and the product

By the definition of a limit, we know that: \(\lim_{x\rightarrow a} f(x) = L\) (given) and we want to prove that \(\lim_{x\rightarrow a} [c f(x)] = cL\) Write the definition of the limit: \(\forall \epsilon>0,\ \exists \delta > 0\) such that for every \(x\) if \(0<|x-a|<\delta\) then \(|f(x)-L|<\epsilon\)

Step 3: Use algebraic manipulations to prove the statement

Using the definition for \(cf(x)\), we can write: \(|c f(x) - cL| = |c||f(x) - L|\) We need to show that for every \(\epsilon>0\), there exists a \(\delta>0\) such that if \(0<|x-a|<\delta\) then \(|c f(x) - cL| < \epsilon\) Now, let \(\epsilon_1 = \frac{\epsilon}{|c|}\), by the definition of limit for \(f(x)\), there exists \(\delta>0\) such that if \(0<|x-a|<\delta\), then \(|f(x)-L|<\epsilon_1\). Therefore, if \(0<|x-a|<\delta\), we have: \(|c f(x) - cL| = |c||f(x) - L| < |c|\epsilon_1\) Since \(|c|\epsilon_1 = \epsilon\), we find that: \(|c f(x) - cL| < \epsilon\) This proves that \(\lim_{x\rightarrow a} [c f(x)] = cL\), thus completing the proof.

Key Concepts

Limit Properties

When studying calculus, understanding the properties of limits is crucial. These properties aid in simplifying the process of finding limits and are valid for all types of functions where the limits exist.

Properties of Limits

For example, one such property states that the limit of a constant times a function equals the constant times the limit of the function. Mathematically, if we have the limit \(\lim_{x\rightarrow a} f(x) = L\), then for any constant \(c\), \(\lim_{x\rightarrow a} [c f(x)] = cL\). This property is particularly useful because it allows us to pull constants out of limits, making our calculations easier.

• Sum limit property: \(\lim_{x\rightarrow a} [f(x) + g(x)] = \lim_{x\rightarrow a} f(x) + \lim_{x\rightarrow a} g(x)\)
• Product limit property: \(\lim_{x\rightarrow a} [f(x) \cdot g(x)] = \lim_{x\rightarrow a} f(x) \cdot \lim_{x\rightarrow a} g(x)\)
• Quotient limit property: Provided \(\lim_{x\rightarrow a} g(x) eq 0\), then \(\lim_{x\rightarrow a} \frac{f(x)}{g(x)} = \frac{\lim_{x\rightarrow a} f(x)}{\lim_{x\rightarrow a} g(x)}\)

Understanding and applying these limit properties can greatly simplify many complex limit calculations, as demonstrated in our exercise example.

Limit Definition

The formal definition of a limit is a foundational concept in calculus, providing a mathematical way to describe how a function behaves as its input approaches a certain value. It's important because it rigorously defines what we mean by saying a function 'approaches' a value.

Understanding the Limit Definition

The formal definition involves two variables: \(\epsilon\) and \(\delta\). \(\epsilon\) is any positive number, no matter how small, and \(\delta\) is a corresponding positive number such that if the difference between \(x\) and an approaching value \(a\) is less than \(\delta\), then the difference between \(f(x)\) and the limit value \(L\) will be less than \(\epsilon\). This definition can be seen in the given solution: for every \(\epsilon > 0\), there exists a \(\delta > 0\) so that if \(0 < |x-a| < \delta\), then \(|f(x) - L| < \epsilon\).
This \(\epsilon-\delta\) approach allows mathematicians to express that the value of \(f(x)\) gets arbitrarily close to \(L\) as \(x\) approaches \(a\), without relying on the notion of 'distance' or 'closeness' but rather with precise inequalities.

Algebraic Manipulation of Limits

Algebraic manipulation of limits involves the use of algebraic techniques to manipulate functions and simplify limits. This is often necessary when a direct substitution leads to an undetermined form or a more complex limit evaluation.

Techniques in Algebraic Manipulation

One common technique, as used in our example, is factoring. By factoring out common terms or using identities, we can simplify the expression involving limits. In the exercise provided, we use algebraic manipulation to prove that \(\lim_{x\rightarrow a} [cf(x)] = cL\). We utilize the absolute value property: \( |cf(x) - cL| = |c||f(x) - L| \).
This relationship makes clear that the 'closeness' of \(cf(x)\) to \(cL\) is directly proportional to the 'closeness' of \(f(x)\) to \(L\), scaled by the absolute value of \(c\). If we can make \(|f(x) - L|\) arbitrarily small (less than some \(\epsilon_1\)), the same can be achieved for \(|cf(x) - cL|\) but with \(\epsilon\).

Using algebraic manipulation in limits can transform expressions into a form where the limit properties can be easily applied. Students are often encouraged to simplify the expressions as much as possible before applying the limit operations to make the problems more tractable.