(a) Prove: If f is continuous at x0 and there are constants...

(a) Prove: If $f$ is continuous at $x_{0}$ and there are constants $a_{0}$ and $a_{1}$ such that $$ \lim _{x \rightarrow x_{0}} \frac{f(x)-a_{0}-a_{1}\left(x-x_{0}\right)}{x-x_{0}}=0 $$ then $a_{0}=f\left(x_{0}\right), f^{\prime}$ is differentiable at $x_{0},$ and $f^{\prime}\left(x_{0}\right)=a_{1} .$ (b) Give a counterexample to the following statement: If $f$ and $f^{\prime}$ are continuous at $x_{0}$ and there are constants $a_{0}, a_{1}$. and $a_{2}$ such that $$ \lim _{x \rightarrow x_{0}} \frac{f(x)-a_{0}-a_{1}\left(x-x_{0}\right)-a_{2}\left(x-x_{0}\right)^{2}}{\left(x-x_{0}\right)^{2}}=0 $$ then $f^{\prime \prime}\left(x_{0}\right)$ exists.

(a) Prove: If $f$ is continuous at $x_{0}$ and there are constants $a_{0}$ and $a_{1}$ such that
$$
\lim _{x \rightarrow x_{0}} \frac{f(x)-a_{0}-a_{1}\left(x-x_{0}\right)}{x-x_{0}}=0
$$
then $a_{0}=f\left(x_{0}\right), f^{\prime}$ is differentiable at $x_{0},$ and $f^{\prime}\left(x_{0}\right)=a_{1} .$
(b) Give a counterexample to the following statement: If $f$ and $f^{\prime}$ are continuous at $x_{0}$ and there are constants $a_{0}, a_{1}$. and $a_{2}$ such that
$$
\lim _{x \rightarrow x_{0}} \frac{f(x)-a_{0}-a_{1}\left(x-x_{0}\right)-a_{2}\left(x-x_{0}\right)^{2}}{\left(x-x_{0}\right)^{2}}=0
$$
then $f^{\prime \prime}\left(x_{0}\right)$ exists.

Key Concepts

Counterexample Method

Constructing a counterexample is a critical method in mathematical analysis to demonstrate the limits and boundaries of a theorem or statement. By showing a specific case where the hypothesis holds but the claimed conclusion fails, one underscores the necessity of additional conditions for the stronger claim, especially in contexts such as the existence of higher-order derivatives.

Continuity

Continuity at a point ensures that the behavior of the function near that point is predictable and that the limit of the function as x approaches the point equals the function’s value at the point. This concept underpins many arguments in calculus, including those concerning differentiability, by ensuring there are no abrupt jumps or gaps in the function's behavior at the point of interest.

Differentiability

Differentiability at a point refers to the existence of a well?defined tangent line to the function at that point, or equivalently, that the function can be locally approximated by a linear function. This is established by the existence of a limit of the difference quotient, making it the bridge between mere continuity and having a derivative.

Limit of the Difference Quotient (Definition of Derivative)

The limit of the difference quotient, formulated as the limit as x approaches the point in question of [f(x) - f(x0)]/(x-x0), is the foundational definition of the derivative. It quantifies the instantaneous rate of change of the function at that point, confirming the linear approximation behavior and ensuring that any suggested coefficient matches the actual derivative when the limit condition is satisfied.

Linear Approximation (First-Order Taylor Approximation)

Linear approximation involves approximating a function near a point using a line, which is characterized by the function’s value and its first derivative at that point. This approximation is crucial in analysis and calculus as it allows for simplified computations and insights into the behavior of functions near specific points, under the assumption of differentiability.

Second-Order Differentiability

Second-order differentiability extends the idea of differentiability to the derivative of the derivative, i.e., the second derivative. This concept is significantly more nuanced than first-order differentiability since the existence of a quadratic approximating term alone does not guarantee that the second derivative exists, highlighting the need for more refined conditions or counterexamples in analysis.

Transcript

00:01 Section 3 .6 problem 111.

00:04 So they give us a piecewise defined function.

00:07 F of x is x sine 1 over x for positive values of x.

00:11 F of x is 0 for values of x less than or equal to 0.

00:15 Show this function is continuous at x equal to 0.

00:20 So the easy case is for negative values leading up to 0, f of x is equal to 0.

00:28 So we know that f of 0 is going to be equal to 0.

00:32 So i have f of 0 is equal to 0.

00:37 If i can show that the limit as x approaches 0 of f of x is equal to 0, then i will have continuity if i can show that.

00:47 So what i'm after, i need to figure out what is the limit.

00:51 I know as i approach from the left, the limit is going to be 0.

00:55 So i need to see what is the limit as x approaches zero from the right of x sine 1 over x.

01:06 If this can be shown to be 0, then i've got continuity of this function.

01:12 So it's just a matter of showing this particular limit.

01:15 So what i know is that the sine value, so i know that the sine of 1 over x is a number between 1 and negative 1.

01:28 Now let's multiply both sides of this equation by x.

01:32 So negative x is less than x, sine 1 over x, which has to be less than x.

01:40 Sorry about this, this was less than or equal to.

01:45 Now, what would happen if you took the limit as x approaches zero of this entire expression? well, what happens on the two extremes, you're going to get zero...

Please give Ace some feedback