Infinite limits of real functions

## Infinite limits of real functions
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            ## Infinite limits
            #### Definition (recall; finite two sided limit)
            $f(x)$ *converges* to *limit* $\ell\in\mathbb R$ as $x\to a$ if

for all $\varepsilon>0$,   $\exists \delta>0$ such that   $0 \lt \lvert x-a \rvert \lt \delta \implies$ $\lvert f(x)-\ell \rvert < \varepsilon$.
 #### Definition (limit $\infty$)
 $f(x)$ *converges* to $\infty$ as $x\to a$ if

Complete this statement!
 
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 ## Infinite limits
 #### Definition (recall; finite two sided limit)
 $f(x)$ *converges* to *limit* $\ell\in\mathbb R$ as $x\to a$ if

for all $M>0$,   $\exists \delta>0$ such that   $0 \lt \lvert x-a \rvert \lt \delta \implies$ $f(x) < -M$.
 
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 ## Infinite limits
 #### Definition (recall; finite two sided limit)
 $f(x)$ *converges* to *limit* $\ell\in\mathbb R$ as $x\to a$ if

for all $M>0$,   $\exists \delta>0$ such that   $0 \lt \lvert x-a \rvert \lt \delta \implies$ $f(x) < -M$.
 #### Exercise
 Write down the definitions of one sided infinite limits (there are four).
 
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 ## Infinite limits
 ### Reminder of some related definitions
 #### Definition (recall; limit from the right)
 $f(x)$ *converges* to *limit* $\ell\in\mathbb R$ as $x\to a^+$ if

for all $\varepsilon>0$,   $\exists \delta>0$ such that   $0 \lt x-a \lt \delta \implies$ $\lvert f(x)-\ell \rvert < \varepsilon$.
 #### Definition (limit $\infty$)
 $f(x)$ *converges* to $\infty$ as $x\to a$ if

for all $M>0$,   $\exists \delta>0$ such that   $0 \lt \lvert x-a \rvert \lt \delta \implies$ $f(x) > M$.
 
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 ## Infinite limits
 ### Reminder of some related definitions
 #### Definition (recall; limit from the right)
 $f(x)$ *converges* to *limit* $\ell\in\mathbb R$ as $x\to a^+$ if

for all $M>0$,   $\exists \delta>0$ such that   $0 \lt \lvert x-a \rvert \lt \delta \implies$ $f(x) > M$.
 #### Solution: Definition (limit $\infty$ from the right)
 $f(x)$ *converges* to $\infty$ as $x\to a^+$ if

for all $M>0$,   $\exists \delta>0$ such that   $0 \lt x-a \lt \delta \implies$ $f(x) > M$.
 
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 ## Example
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Plot $y=f(x)$

![Desmos plot of y=1 on 1000 x minus 1 cubed](gfx/1on1000xminus1cubed.png "Desmos plot of y=1/[1000(x-1)^3]")

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$\displaystyle f(x) = \frac1{1000(x-1)^3}$.
              #### Theorem
              $\displaystyle\lim\_{x\to1^-} f(x) = -\infty$ and $\displaystyle\lim\_{x\to1^+} f(x) = \infty$, so $\displaystyle\lim\_{x\to1} f(x)$ is undefined.

#### Proof
              Try it!

#### Definition (limit $\infty$ from the right)
 $f(x)$ *converges* to $\infty$ as $x\to a^+$ if
 for all $M>0$,   $\exists \delta>0$ such that   $0 \lt x-a \lt \delta \implies$ $f(x) > M$.
 #### How to prove convergence if you know the limit
 Construct a machine that tells you $\delta$ for any given $M$, so $\delta$ does the job.
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 ## Example
 <div style="display:flex;flex-direction:row;width:100%">
 <div style="flex:1 1 0px;margin-right:2vw;background-color:white;color:black">

Plot $y=f(x)$

![Desmos plot of y=1 on 1000 x minus 1 cubed](gfx/1on1000xminus1cubed.png "Desmos plot of y=1/[1000(x-1)^3]")

</div>
 <div style="flex:4.5 4.5 0px;margin-left:2vw">

#### Proof
              For any $M>0$, let $\delta=\frac1{10\sqrt[3]M}$.

If $0<x-1<\delta$, then $\frac1{x-1} \gt 10 \sqrt[3]M$, so $\frac1{10(x-1)} \gt \sqrt[3]M$, so $f(x) = \left(\frac1{10(x-1)}\right)^3 \gt M$.
 Therefore $\displaystyle\lim\_{x\to1^+} f(x) = \infty$.

If instead $0<1-x<\delta$, then $\frac1{x-1} \lt - 10 \sqrt[3]M$, so $\frac1{10(x-1)} \lt -\sqrt[3]M$, so $f(x) = \left(\frac1{10(x-1)}\right)^3 \lt -M$.
 Hence $\displaystyle\lim\_{x\to1^-} f(x) = -\infty$.
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