MA5131 Summary 1
Limits
Intuitively, $\lim\limits_{x \to k} f(x)=v$ means that for some interval around $x$, but not including $x$, the values of $f$ in that interval is around $v$.
For continuous functions, $\lim\limits_{x \to k} f(x)=f(k)$. Such functions includes polynomials, exponentials and some trigonometric functions ($\sin$ and $\cos$).
Limits can be broken up if both limits exists for $+-\times \div$. That is $\lim\limits_{x \to k} (f(x) \otimes g(x))=\lim\limits_{x \to k} f(x) \otimes \lim\limits_{x \to k} g(x)$, where $\otimes$ is one of $+-\times \div$. Note that both limits must exist and be well-defined. Furthermore, for division $\lim\limits_{x \to k} g(x) \ne 0$.
Sometimes, you will have to evaluate limits of the form $\lim\limits_{x \to k} \frac{f(x)}{g(x)}$ where $\lim\limits_{x \to k} f(x)=0$ and $\lim\limits_{x \to k} g(x)=0$. Since L’Hopital’s rule is not allowed in this chapter, the limit is usually some variant of $\lim\limits_{x \to k} \frac{a^2-b^2}{a-b}$ or $\lim\limits_{x \to 0} \frac{\sin x}{x}$.
$\lim\limits_{x \to 3} \frac{x-3}{\sqrt{x+1}-2}=\lim\limits_{x \to 3} \frac{(\sqrt{x+1}+2)(\sqrt{x+1}-2)}{\sqrt{x+1}-2}=\lim\limits_{x \to 3} \sqrt{x+1}+2=4$
$\lim\limits_{x \to 0} \frac{\cos (x)-1}{\sin^2(x)}=\lim\limits_{x \to 0} \frac{\cos(x)-1}{1-\cos^2(x)}=\lim\limits_{x \to 0} \frac{\cos(x)-1}{(1 - \cos (x))(1 + \cos (x))}=-\lim\limits_{x \to 0} \frac{1}{1+ \cos (x)}=-\frac 12$
$\lim\limits_{x \to 0} \frac{\sin (cx)}{x}=\lim\limits_{x \to 0} \frac{\sin (cx)}{cx} \cdot \lim\limits_{x \to 0} c=1 \cdot c=c$
Note that if a function is only defined at a single point $(c,f(c))$, then $\lim\limits_{x \to c} f(x)$ is undefined. This is because there are not points on a open interval around $c$, not including $c$, that is defined.
Furthermore, even if both $\lim\limits_{x \to k} f(x)$ and $f(k)$ are defined. It is not true that $\lim\limits_{x \to k} f(x)=f(k)$.
One-Sided Limits
Given function $f(x)=\begin{cases} a(x), &\text{if } x < k \\ b(x), &\text{if } x = k \\ c(x), &\text{if } x > k\end{cases}$
We say $\lim\limits_{x \to k^-}f(x)=\lim\limits_{x \to k^-} a(x)$ and $\lim\limits_{x \to k^+}f(x)=\lim\limits_{x \to k^+} c(x)$ in this case.
If $\lim\limits_{x \to k^-}f(x)=\lim\limits_{x \to k^+}f(x)=L$, we say $\lim\limits_{x \to k} f(x)=L$. Else we say that $\lim\limits_{x \to k} f(x)$ is undefined.
Notice that in all $3$ limits, the value of $b(k)$ is not used. When evaluating a limit towards $k$, you don’t have to care about the actual value of $b(k)$.
Note that $\lvert x \rvert =\begin{cases} -x, &\text{if } x < 0 \\ 0, &\text{if } x = 0 \\ x, &\text{if } x > 0\end{cases}$
$\lim\limits_{x \to 3^-} \frac{\lvert x-1 \rvert-\lvert x-5 \rvert}{\lvert x^2 -2x -3 \rvert}=\lim\limits_{x \to 3^-} \frac{(x-1)-(5-x)}{-(x^2-2x-3)}=\lim\limits_{x \to 3^-} -\frac{(x-1)-(5-x)}{x^2-2x-3}=\lim\limits_{x \to 3^-} -\frac{2x-6}{(x-3)(x+1)}=\lim\limits_{x \to 3^-} -\frac{2}{(x+1)}=-\frac{2}{4}=-\frac 12$
$\lim\limits_{x \to 3^+} \frac{\lvert x-1 \rvert- \lvert x-5 \rvert}{\lvert x^2 -2x -3 \rvert}=\lim\limits_{x \to 3^+} \frac{(x-1)-(5-x)}{x^2-2x-3}=\lim\limits_{x \to 3^+} \frac{(x-1)-(5-x)}{x^2-2x-3}=\lim\limits_{x \to 3^+} \frac{2x-6}{(x-3)(x+1)}=\lim\limits_{x \to 3^+} \frac{2}{(x+1)}=\frac{2}{4}=\frac 12$
Squeeze Theorem
Suppose $a(x) \leq b(x) \leq c(x) \,\forall\, x \in I \setminus {k}$, for some open interval $I$ containing $k$. Then if $\lim\limits_{x \to k} a(x) = \lim\limits_{x \to k} c(x)=L$, then $\lim\limits_{x \to k} b(x)=L$.
An important proof using this theorem is that $\lim\limits_{x \to 0} \frac{\sin x}{x}=1$. We can show that $\frac{\sin(x)}{2} \leq \frac x2 \leq \frac {\tan x}{2}$. By focusing on the $x > 0$ case, we multiply all sides by by $\frac 2 {\sin (x)}$, we obtain $1 \leq \frac{x}{\sin (x)} \leq \frac{1}{\cos (x)}$, since $\frac{2}{\sin (x)}>0$ for $0 < x < \frac \pi 2$. We can show that the inequality holds for $-\frac{\pi}{2} < x <0$ with a similar argument.
Since $\lim\limits_{x \to 0} 1=1$ and $\lim\limits_{x \to 0} \frac 1 {\cos (x)}=1$, we can show that $\lim\limits_{x \to 0}\frac{x}{\sin (x)}=1$. Thus, $\lim\limits_{x \to 0} \frac{\sin (x)}{x} = \frac{\lim\limits_{x \to 0} 1}{\lim\limits_{x \to 0} \frac{x}{\sin (x)}}=\frac{1}{1}=1$.
Suppose we want to find $\lim\limits_{x \to 0} x \cdot f(x)$ and we know that $\lvert f(x) \rvert \leq \lvert g(x) \rvert \,\forall\, x \in \mathbb{R}$. Since $-\lvert x \rvert \leq x \leq \lvert x \rvert, \,\forall\, x \in \mathbb{R}$, we can show that $-\lvert x \cdot g(x) \lvert \leq x \cdot f(x) \leq \lvert x \cdot g(x) \rvert$. If we chose a good $g$ (usually $g(x)=c$, for some constant $c$) such that $\lim\limits_{x \to 0} \lvert x \cdot g(x) \rvert=0$ can be easily shown, then we can show that $\lim\limits_{x \to 0} x \cdot f(x)=0$.
We can do a similar thing for finding $\lim\limits_{x \to 0} x^2 \cdot f(x)$. This is even simpler as $- x^2 \leq x^2 \leq x^2$ since $x^2 \geq 0$.
For squeeze theorem you can sometimes be very lazy with inequalities. There is no reason to write $-1 \leq \sin (x) \leq 1$ when you can write $-10^{100} \leq \sin (x) \leq 10^{100}$ which may suffice and may be an easier bound to show, especially for more complicated functions.
Continuity
A point $c$ is said to be continuous on $f$ if $\lim\limits_{x \to c} f(x)=f(c)$. As a direct consequence, both $\lim\limits_{x \to c} f(x)$ and $f(c)$ have to exist and be well-defined (not infinity). Intuitively this means that $(c,f(c))$ exists and the points around it are near it.
An interval $(l,r)$ is said to be continuous on $f$ if for all $c \in (l,r)$, $c$ is continuous on $f$. For interval $[l,r]$ to be continuous, $(l,r)$ must be continuous and $\lim\limits_{x \to l^+}f(x)=f(l)$ and $\lim\limits_{x \to r^-}f(x)=f(r)$.
If $f$ is continuous and $f^{-1}$ exists, then $f^{-1}$ is continuous.
If $f$ and $g$ are continuous at $c$, then $f \pm g$, $fg$ and $\frac fg$ ($g \ne 0$) are all continuous at $c$.
Corollary 1: all functions of the form $\frac{p_1(x)}{p_2(x)}$ where $p_1$ and $p_2$ are polynomials are continuous everywhere except $c$ such that $p_2(c)=0$.
Intermediate Value Theorem
Suppose $f$ is continuous on $[l,r]$, then for any $k \in [\min(f(l),f(r)), \max(f(l),f(r))]$, there exists $c \in [l,r]$ such that $f(c)=k$. There exist another version of this theorem that states for any $k \in (\min(f(l),f(r)), \max(f(l),f(r)))$, there exists $c \in (l,r)$ such that $f(c)=k$.
Corollary 1: Suppose $f$ is continuous on $[l,r]$ and $f(l) \cdot f(r) \leq 0$, then there exists $c \in [l,r]$ such that $f(c)=0$. Alternatively, if $f(l) \cdot f(r) < 0$, then there exists $c \in (l,r)$ such that $f(c)=0$.
Corollary 2: Suppose both $f$ and $g$ are continuous on $[l,r]$ and $f(l) \leq g(l)$ and $f(r) \geq g(r)$, then there exists $c \in [l,r]$ such that $f(c)=g(c)$.
Proof: Define $h(x)=f(x)-g(x)$. Then, $h(l) \geq 0$ and $h(r) \leq 0$. Thus, $h(l) \cdot h(r) \leq 0$. Since $h$ is also continuous on $[l,r]$, by intermediate value theorem, there exists $c\in [l,r]$ such that $h(c)=f(c)-g(c)=0$, or that $f(c)=g(c)$.
Extreme Value Theorem
Suppose $f$ is continuous on $[l,r]$, then there exists a minimum and a maximum value on $[l,r]$.
Note that if $f$ is only continuous on $(l,r)$, there may not exists a minimum nor maximum value on $[l,r]$.
Counterexample: $f(x)=\begin{cases} 2, &\text{if } x =0 \\ x, &\text{if } 0 < x \leq 2\end{cases}$
In this graph, there is no minimum value on $[0,2]$ despite $(0,2)$ being continuous on $f$. Suppose $c \in [0,2]$, if $c=0$, it is not minimum as $f(1)<f(0)$. If $c \ne 0$, $f(\frac c2)<f(c)$. Therefore, there is no absolute minimum on $[0,2]$.