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        <div toc=""><ul toc=""><li entry="h1"><a href="#chapter-1">Chapter 1</a></li><li entry="h2"><a href="#%C2%A71.5-%7C%20the%20limit%20of%20a%20function">§1.5 | The Limit of a Function</a></li><li entry="h3"><a href="#vertical-asymptote-def">Vertical Asymptote (Definition §1.5.6)</a></li><li entry="h2"><a href="#%C2%A71.6-%7C%20calculating%20limits%20using%20the%20limit%20laws">§1.6 | Calculating Limits Using the Limit Laws</a></li><li entry="h3"><a href="#limit-laws">Limit Laws</a></li><li entry="h3"><a href="#power-law">Power Law</a></li><li entry="h3"><a href="#root-law">Root Law</a></li><li entry="h2"><a href="#%C2%A71.8-%7C%20continuity">§1.8 | Continuity</a></li><li entry="h3"><a href="#the-intermediate%20value%20theorem%20(ivt)">The Intermediate Value Theorem (IVT)</a></li></ul></div>
    </header>
    <h1 id="chapter-1">Chapter 1</h1>

<img style="max-width: 800px;" src="https://cdn.kastatic.org/ka-perseus-images/a193121287b64721de28fcbfaec9f5919a367dd2.png">
<h2 id="%C2%A71.5-%7C%20the%20limit%20of%20a%20function">§1.5 | The Limit of a Function</h2>

<section class="note-block block"><h3 id="vertical-asymptote-def">Vertical Asymptote (Definition §1.5.6)</h3><p>
    
    The vertical line <span inline="" math-inline="">\(x = a\)</span> is called the <b>vertical asymptote</b> of the curve <span inline="" math-inline="">\(y = f(x)\)</span> if at least one of the following statements is true:
    
    </p><div block="" math-block="">$$ \begin{equation}
\begin{split}
    \lim_{x \to a^{-}} f(x) &= \infty \\
    \lim_{x \to a^{-}} f(x) &= -\infty \\
    \lim_{x \to a^{+}} f(x) &= \infty \\
    \lim_{x \to a^{+}} f(x) &= -\infty
\end{split}
\end{equation} $$</div><p>


    Also:
    </p><div block="" math-block="">$$\begin{equation}
\begin{split}
\lim_{x \to a} f(x) &= \lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x) = \infty \\
\lim_{x \to a} f(x) &= \lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x) = -\infty
\end{split}
\end{equation}$$</div><p>

    To my understanding, if there is a vertical asymptote at <span inline="" math-inline="">\(x = a\)</span>, then <span inline="" math-inline="">\(\lim_{x \to a}f(x)\)</span> is technically undefined. But instead, of saying undefined, we say <span inline="" math-inline="">\(\pm \infty\)</span> since it gives more information. I.e. this is a special case of being undefined.
</p></section>


<h2 id="%C2%A71.6-%7C%20calculating%20limits%20using%20the%20limit%20laws">§1.6 | Calculating Limits Using the Limit Laws</h2>

<section class="note-block block"><h3 id="limit-laws">Limit Laws</h3><p>

    Suppose that <span inline="" math-inline="">\(c\)</span> is a constant and that the limits <span inline="" math-inline="">\(\lim_{x \to a} f(x)\)</span> and <span inline="" math-inline="">\(\lim_{x \to a} g(x)\)</span> exists, then:
    
    </p><div block="" math-block="">$$ \begin{equation}
\begin{split}
    \lim_{x \to a} \Big[ f(x) + g(x) \Big] &= \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \\
    \lim_{x \to a} \Big[ f(x) - g(x) \Big] &= \lim_{x \to a} f(x) - \lim_{x \to a} g(x) \\
    \lim_{x \to a} \Big[ c \cdot f(x) \Big] &= c \cdot \lim_{x \to a} f(x) \\
    \lim_{x \to a} \Big[ f(x) \cdot g(x) \Big] &= \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \\
    \lim_{x \to a} \frac{f(x)}{g(x)} &= \frac{\lim_{x \to a} f(x)}{\lim_{x \to a}g(x)} \; \text{if} \lim_{x \to a}g(x) \neq 0
\end{split}
\end{equation} $$</div></section>

<section class="note-block block"><h3 id="power-law">Power Law</h3><div block="" math-block="">$$\lim_{x \to a} \Big[ f(x) \Big]^n = \Big[ \lim_{x \to a} f(x) \Big]^n$$</div><section warning="" class="note-block block"><p><span inline="" math-inline="">\(n\)</span> must be a positive integer.</p></section></section>

<section class="note-block block"><h3 id="root-law">Root Law</h3><div block="" math-block="">$$\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to a} f(x)}$$</div><section warning="" class="note-block block"><p><span inline="" math-inline="">\(n\)</span> must be a positive integer.
    </p></section><section class="note-block block"><p>
        If <span inline="" math-inline="">\(n\)</span> is even, we assume that <span inline="" math-inline="">\(\lim_{x \to a} f(x) > 0\)</span>.
    </p></section></section>

<h2 id="%C2%A71.8-%7C%20continuity">§1.8 | Continuity</h2>

<section class="note-block block"><h3 id="the-intermediate%20value%20theorem%20(ivt)">The Intermediate Value Theorem (IVT)</h3><p>

    Suppose that <span inline="" math-inline="">\(f\)</span> is continuous on the closed interval <span inline="" math-inline="">\([a, b]\)</span>, and let N be any number between <span inline="" math-inline="">\(f(a)\)</span> and <span inline="" math-inline="">\(f(b)\)</span>, where <span inline="" math-inline="">\(f(a) \neq f(b)\)</span>, then there exists a number <span inline="" math-inline="">\(c\)</span> in <span inline="" math-inline="">\((a, b)\)</span> such that <span inline="" math-inline="">\(f(c) = N\)</span>.

    </p><p>Or alternatively, suppose <span inline="" math-inline="">\(f\)</span> is a function that is continuous at every point in the interval <span inline="" math-inline="">\([a, b]:\)</span></p><ul>
    <li><span inline="" math-inline="">\(f\)</span> will take on every value between <span inline="" math-inline="">\(f(a)\)</span> and <span inline="" math-inline="">\(f(b)\)</span> over the interval.</li>
    <li>For any L between the values <span inline="" math-inline="">\(f(a)\)</span> and <span inline="" math-inline="">\(f(b)\)</span>, there exists a number <span inline="" math-inline="">\(c\)</span> in <span inline="" math-inline="">\([a, b]\)</span> for which <span inline="" math-inline="">\(f(c) = L\)</span>.</li>
</ul><section class="note-block block"><ul>
        <li>The intermediate value theorem states that if a continuous function attains two values, it must also attain all values in between these two values.</li>
        <li>The Intermediate Value Theorem states that a continuous function takes on every intermediate value between the function values <span inline="" math-inline="">\(f(a)\)</span> and <span inline="" math-inline="">\(f(b)\)</span>.</li>
        <li>The statement of the theorem has multiple requirements, all of which are necessary for the conclusion to hold.</li>
    </ul></section><section class="note-block block"><p>
    The IVT can be used to show that a root exists, if given some interval between <span inline="" math-inline="">\(a\)</span> and <span inline="" math-inline="">\(b\)</span>: 

    </p><ul>
        <li><span inline="" math-inline="">\(f(a)\)</span> is negative</li>
        <li><span inline="" math-inline="">\(f(b)\)</span> is positive</li>
    </ul><p>

    or

    </p><ul>
        <li><span inline="" math-inline="">\(f(a)\)</span> is positive</li>
        <li><span inline="" math-inline="">\(f(b)\)</span> is negative</li>
    </ul><p>then therefore, if the function is continuous, then there must be a value for which some <span inline="" math-inline="">\(x\)</span> in <span inline="" math-inline="">\(f(x) = 0\)</span> exists. I.e. that a root exists. Which should make sense.</p><p>**More generally**, this can be extended to any value between some interval. For instance, if:</p><ul>
        <li><span inline="" math-inline="">\(f(a) = 3\)</span></li><span inline="" math-inline="">\(f(a) = 9\)</span>
    </ul><p>Then there must exist some value that we will call <span inline="" math-inline="">\(x\)</span>, for which:</p><ul>
        <li><span inline="" math-inline="">\(f(x) = 6\)</span></li>
    </ul></section></section>
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