Question

Creating functions satisfying given limit conditions Find a function \(f\) satisfying \(\lim _{x \rightarrow 1} \frac{f(x)}{x-1}=2\).

Short answer

Question: Find a function \(f(x)\) such that \(\lim_{x\rightarrow 1} \frac{f(x)}{x-1} = 2\). Answer: One possible function is \(f(x) = a(x-1)^2 + (x-1) + 1\), where \(a\) is any constant.

Step-by-step solution

Choose a Polynomial Function

We will look for a polynomial function that satisfies the given limit condition. Since the denominator is \(x-1\), a good starting point is to choose \(f(x) = a(x-1)^2 + b(x-1) + c\), where \(a\), \(b\), and \(c\) are constants. This will allow us to simplify the fraction when taking the limit.

Set up the Limit Expression

With the chosen polynomial function, let's set up the limit expression: $$\lim _{x \rightarrow 1} \frac{a(x-1)^2 + b(x-1) + c}{x-1}$$

Simplify the Limit Expression

To simplify, we can factor out \((x-1)\) from the numerator: $$\lim _{x \rightarrow 1} \frac{(x-1)(a(x-1) + b) + c}{x-1}$$ Now, as \(x \rightarrow 1\), the terms \((x-1)\) in both the numerator and the denominator cancel out and we have: $$\lim _{x \rightarrow 1} (a(x-1) + b) + c$$

Evaluate the Limit

When \(x \rightarrow 1\), we can substitute the value of \(x\) into the expression to find the limit: $$\lim _{x \rightarrow 1} (a(1-1) + b) + c = b + c$$

Compare with Given Condition

Our goal is to find a function such that the computed limit is equal to 2: $$b + c = 2$$

Choose Values for Constants

We can choose any values for constants \(b\) and \(c\) such that their sum is 2, and we will have a valid function \(f(x)\). For example, we can choose \(b=1\) and \(c=1\).

Construct the Function

With the chosen constants \(b=1\), \(c=1\), and \(a\) being a free parameter, our function \(f(x)\) is: $$f(x) = a(x-1)^2 + (x-1) + 1$$ This function satisfies the given limit condition, and we are free to choose any value for \(a\).

Key concepts

Polynomial function

A polynomial function is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. For example, in this exercise, we explored a polynomial of the form

• \( f(x) = a(x-1)^2 + b(x-1) + c \)

This polynomial is composed of terms with different powers of \((x-1)\), making it suitable for our needs.
Choosing a polynomial function helps us manipulate the expression more easily when solving limits.
This is because polynomials are continuous and have simple properties that make calculation straightforward.

Evaluating limits

Evaluating limits involves finding the value that a function approaches as the input approaches a certain point.
In this problem, we needed to evaluate the limit \( \lim_{x \rightarrow 1} \frac{f(x)}{x-1} \).
Limits are foundational in calculus and describe behavior at specific points which might not be explicitly defined in the original function.When dealing with limits, especially involving division, we often face indeterminate forms such as \( \frac{0}{0} \).
This demands additional steps like factoring and simplification to resolve the expression.

Limit expression simplification

Limit expression simplification is the process of breaking down complex fractions or expressions into easier ones to evaluate.
In our exercise, we started by factoring out \((x-1)\) from the numerator:

• \( \lim_{x \rightarrow 1} \frac{(x-1)(a(x-1) + b) + c}{x-1} \)

By canceling out \((x-1)\), we managed to simplify the expression to \((a(x-1) + b) + c \).
This makes the limit straightforward as we can directly substitute \(x=1\) to evaluate the expression.

Polynomial limit conditions

Polynomial limit conditions help us create functions that behave in a specific desired manner as a variable approaches a given point.
For this exercise's condition, we needed \(b+c=2\).
This condition guided us in choosing the appropriate constants for our polynomial.
Once the condition is known, selecting values for the constants becomes a balancing act to satisfy the original limit requirement.
In our example, choosing \(b=1\) and \(c=1\) provides a simple solution that fits perfectly.