Problem 13 Suppose that $f$ is differenti... [FREE SOLUTION]

Chapter 1: Problem 13

Suppose that $f$ is differentiable at $x_{0}$. Let $L$ be the "best linear approximation" defined by $L(x)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)(x-$ $\left.x_{0}\right)$. Given that $R(x)=f(x)-L(x)$, show that $$ \lim _{x \rightarrow x_{0}} \frac{R(x)}{x-x_{0}}=0 $$

Short Answer

Expert verified

The limit is 0.

Step by step solution

Understand the Problem Statement

We are given a function $ f $ that is differentiable at $ x_0 $. The task is to find the limit of the ratio $ \frac{R(x)}{x-x_0} $ as $ x $ approaches $ x_0 $, where $ R(x) = f(x) - L(x) $ and $ L(x) = f(x_0) + f'(x_0)(x-x_0) $.

Decompose the Remainder Function R(x)

Start by substituting the definition of $ L(x) $ into $ R(x) $:\[ R(x) = f(x) - \left(f(x_0) + f'(x_0)(x-x_0)\right). \] Simplifying this gives: \[ R(x) = f(x) - f(x_0) - f'(x_0)(x-x_0). \]

Apply the Definition of Derivative

Since $ f $ is differentiable at $ x_0 $, the definition of the derivative tells us that: \[ \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} = f'(x_0). \] Rearrange the expression for $ f(x) - f(x_0) $:\[ f(x) - f(x_0) = f'(x_0)(x-x_0) + \varepsilon(x) \cdot (x-x_0), \] where $ \varepsilon(x) \to 0 $ as $ x \to x_0 $.

Substitute Back into the Remainder Expression

Substitute the expression for $ f(x) - f(x_0) $ into $ R(x) $:\[ R(x) = f'(x_0)(x-x_0) + \varepsilon(x) \cdot (x-x_0) - f'(x_0)(x-x_0). \] This simplifies to \[ R(x) = \varepsilon(x) \cdot (x-x_0). \]

Find the Limit of R(x)/(x-x_0)

Compute the limit: \[ \lim_{x \to x_0} \frac{R(x)}{x-x_0} = \lim_{x \to x_0} \frac{\varepsilon(x) \cdot (x-x_0)}{x-x_0} = \lim_{x \to x_0} \varepsilon(x). \] Since $ \varepsilon(x) \to 0 $ as $ x \to x_0 $, it follows that: \[ \lim_{x \to x_0} \frac{R(x)}{x-x_0} = 0. \]

Conclusion

By showing that $ \varepsilon(x) \to 0 $ as $ x \to x_0 $, and using the expression for $ R(x) $, we confirmed that $ \lim_{x \to x_0} \frac{R(x)}{x-x_0} = 0 $, as required by the problem statement.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Approximation

Linear approximation is a method used to estimate the value of a function near a given point by using its tangent line. It provides a way to approximate complex functions with simpler linear functions that are easy to compute and understand.
For a differentiable function, the linear approximation at a point $x_0$ is given by the equation $L(x) = f(x_0) + f'(x_0)(x - x_0)$. This equation represents the tangent line to the graph of the function at $x_0$.

$f(x_0)$ is the value of the function at the point $x_0$.
$f'(x_0)$ is the slope of the tangent line, which is the derivative of the function at $x_0$.
$x - x_0$ represents the horizontal distance from $x_0$.

This approximation is most accurate near $x_0$, and accuracy decreases as you move away from this point. It's especially useful in calculus for simplifying calculations and analyzing the behavior of functions.

Differentiable Function

A function is called differentiable at a point if it has a well-defined tangent at that point, meaning its derivative exists there. Differentiability implies that the function is smooth and continuous around that specific point.
If a function $f$ is differentiable at $x_0$, it means that the limit
\[\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} \]
exists and equals $f'(x_0)$, the derivative of $f$ at $x_0$. This derivative represents the slope of the tangent line at the point $x_0$.

Differentiability assures that small changes in $x$ cause small changes in $f(x)$, which is a crucial property for analyzing functions using calculus tools.
If a function is differentiable at a point, it is also continuous at that point, but the reverse is not necessarily true.

The concept of differentiability is foundational in differential calculus, as it allows us to use linear approximations to model and predict function behavior around specific points.

Limit of a Function

The limit of a function at a certain point is a fundamental concept in calculus. It describes the value that a function approaches as the input approaches a specific point.
In the problem, we examined the limit$ \lim _{x \rightarrow x_{0}} \frac{R(x)}{x-x_{0}}$, which shows how the difference between the actual function $f(x)$ and its linear approximation $L(x)$ behaves near $x_0$.

The remainder $R(x)$, defined as $f(x) - L(x)$, demonstrates how much the actual function value deviates from the linear approximation.
We determined that this remainder divided by $x - x_0$ tends to zero as $x$ approaches $x_0$, meaning that $R(x)$ gets arbitrarily small compared to $x - x_0$.

Understanding limits allows students to handle cases where direct computation is challenging, by revealing underlying patterns and behaviors of functions.

91影视

Suppose that \(f\) is differentiable at \(x_{0}\). Let \(L\) be the "best linear approximation" defined by \(L(x)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)(x-\) \(\left.x_{0}\right)\). Given that \(R(x)=f(x)-L(x)\), show that $$ \lim _{x \rightarrow x_{0}} \frac{R(x)}{x-x_{0}}=0 $$

Short Answer

Step by step solution

Understand the Problem Statement

Decompose the Remainder Function R(x)

Apply the Definition of Derivative

Substitute Back into the Remainder Expression

Find the Limit of R(x)/(x-x_0)

Conclusion

Key Concepts

Linear Approximation

Differentiable Function

Limit of a Function

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