$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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Differentiable Functions are Continuous

Let $A\subseteq\mathbb{R}^n$, $\vec{a}\in \text{int}(A)$, and $f:A\to\mathbb{R}^m$. If $f$ is differentiable at $\vec{A}$, then it is continuous at $\vec{a}$.

Concepts

If a function is differentiable at a point, then it has to be continuous at a point.
this seems to only apply for single variable functions. Multivariable, you can have a function that's not continuous at a point but the derivative still existing. Is it because of the different limit needing to exist?

Hypothesis

The function needs to be differentiable, as the one-variable limit definition of derivatives is used in the proof.

Results

Continuity leads to A Lot.

Proof

To prove that it is continuous at a point, we need the limit of $f(a+h)-f(a)$ to go to $0$ as $h$ goes to $0$. Notice how this is the numerator in the definition of the derivative.
Let $f$ be a function, and let $f$ be differentiable at $a$. Then for any $h\neq= 0$, we have $$\begin{align*} f(a+h)-f(a)=h\cdot\frac{f(a+h)-f(a)}{h} \end{align*}$$. Taking the limit as $h$ goes to $0$, we get $$\begin{align*} \lim_{h\to 0}[f(a+h)-f(a)]&=\lim_{h\to 0}\times \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\\ &=0\times f'(a) &=0, \end{align*}$$ and hence $$\begin{align*} \lim_{h\to 0}f (a+h)=f(a) \end{align*}$$ Let $x=a+h$. Then $\lim_{h\to 0}f(a+h)=\lim_{x\to a}f(x)$ giving $$\begin{align*} \lim_{h\to 0} f(a+h)&=f(a)\\ \lim_{x\to a}f(x)&=f(a) \end{align*}$$. Hence \lim_{x\to a}f(x) = f(a) means the function the continuous at $a$.

Differentiable Functions are Continuous

Let $A\subseteq\mathbb{R}^n$, $\vec{a}\in \text{int}(A)$, and $f:A\to\mathbb{R}^m$. If $f$ is differentiable at $\vec{A}$, then it is continuous at $\vec{a}$.

Concepts

If a function is differentiable at a point, then it has to be continuous at a point.
this seems to only apply for single variable functions. Multivariable, you can have a function that's not continuous at a point but the derivative still existing. Is it because of the different limit needing to exist?

Hypothesis

The function needs to be differentiable, as the one-variable limit definition of derivatives is used in the proof.

Results

Continuity leads to A Lot.

Proof

To prove that it is continuous at a point, we need the limit of $f(a+h)-f(a)$ to go to $0$ as $h$ goes to $0$. Notice how this is the numerator in the definition of the derivative.
Let $f$ be a function, and let $f$ be differentiable at $a$. Then for any $h\neq= 0$, we have $$\begin{align*} f(a+h)-f(a)=h\cdot\frac{f(a+h)-f(a)}{h} \end{align*}$$. Taking the limit as $h$ goes to $0$, we get $$\begin{align*} \lim_{h\to 0}[f(a+h)-f(a)]&=\lim_{h\to 0}\times \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\\ &=0\times f'(a) &=0, \end{align*}$$ and hence $$\begin{align*} \lim_{h\to 0}f (a+h)=f(a) \end{align*}$$ Let $x=a+h$. Then $\lim_{h\to 0}f(a+h)=\lim_{x\to a}f(x)$ giving $$\begin{align*} \lim_{h\to 0} f(a+h)&=f(a)\\ \lim_{x\to a}f(x)&=f(a) \end{align*}$$. Hence \lim_{x\to a}f(x) = f(a) means the function the continuous at $a$.
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FullPage
result
concepts
hypothesis
implications
proof

Differentiable Functions are Continuous

Let $A\subseteq\mathbb{R}^n$, $\vec{a}\in \text{int}(A)$, and $f:A\to\mathbb{R}^m$. If $f$ is differentiable at $\vec{A}$, then it is continuous at $\vec{a}$.

Concepts

If a function is differentiable at a point, then it has to be continuous at a point.
this seems to only apply for single variable functions. Multivariable, you can have a function that's not continuous at a point but the derivative still existing. Is it because of the different limit needing to exist?

Hypothesis

The function needs to be differentiable, as the one-variable limit definition of derivatives is used in the proof.

Results

Continuity leads to A Lot.

Proof

To prove that it is continuous at a point, we need the limit of $f(a+h)-f(a)$ to go to $0$ as $h$ goes to $0$. Notice how this is the numerator in the definition of the derivative.
Let $f$ be a function, and let $f$ be differentiable at $a$. Then for any $h\neq= 0$, we have $$\begin{align*} f(a+h)-f(a)=h\cdot\frac{f(a+h)-f(a)}{h} \end{align*}$$. Taking the limit as $h$ goes to $0$, we get $$\begin{align*} \lim_{h\to 0}[f(a+h)-f(a)]&=\lim_{h\to 0}\times \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\\ &=0\times f'(a) &=0, \end{align*}$$ and hence $$\begin{align*} \lim_{h\to 0}f (a+h)=f(a) \end{align*}$$ Let $x=a+h$. Then $\lim_{h\to 0}f(a+h)=\lim_{x\to a}f(x)$ giving $$\begin{align*} \lim_{h\to 0} f(a+h)&=f(a)\\ \lim_{x\to a}f(x)&=f(a) \end{align*}$$. Hence \lim_{x\to a}f(x) = f(a) means the function the continuous at $a$.

Differentiable Functions are Continuous

Let $A\subseteq\mathbb{R}^n$, $\vec{a}\in \text{int}(A)$, and $f:A\to\mathbb{R}^m$. If $f$ is differentiable at $\vec{A}$, then it is continuous at $\vec{a}$.

Concepts

If a function is differentiable at a point, then it has to be continuous at a point.
this seems to only apply for single variable functions. Multivariable, you can have a function that's not continuous at a point but the derivative still existing. Is it because of the different limit needing to exist?

Hypothesis

The function needs to be differentiable, as the one-variable limit definition of derivatives is used in the proof.

Results

Continuity leads to A Lot.

Proof

To prove that it is continuous at a point, we need the limit of $f(a+h)-f(a)$ to go to $0$ as $h$ goes to $0$. Notice how this is the numerator in the definition of the derivative.
Let $f$ be a function, and let $f$ be differentiable at $a$. Then for any $h\neq= 0$, we have $$\begin{align*} f(a+h)-f(a)=h\cdot\frac{f(a+h)-f(a)}{h} \end{align*}$$. Taking the limit as $h$ goes to $0$, we get $$\begin{align*} \lim_{h\to 0}[f(a+h)-f(a)]&=\lim_{h\to 0}\times \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\\ &=0\times f'(a) &=0, \end{align*}$$ and hence $$\begin{align*} \lim_{h\to 0}f (a+h)=f(a) \end{align*}$$ Let $x=a+h$. Then $\lim_{h\to 0}f(a+h)=\lim_{x\to a}f(x)$ giving $$\begin{align*} \lim_{h\to 0} f(a+h)&=f(a)\\ \lim_{x\to a}f(x)&=f(a) \end{align*}$$. Hence \lim_{x\to a}f(x) = f(a) means the function the continuous at $a$.
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result
concepts
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proof