Limits of real functions

## Calculating limits of real functions
            ---
            ## Calculating limits of real functions
            #### Definition
            Suppose $c\lt a\lt d$ and $f:(c,a)\cup(a,d):\to\mathbb R$.
            Then

$f(x)$ *converges* to *limit* $\ell\in\mathbb R$ as $x\to a$ if

<span class="color1">for all $\varepsilon>0$</span>,   <span class="color6">$\exists \delta>0$ such that</span>   <span class="color5">$0 \lt \lvert x-a \rvert \lt \delta \implies$</span> <span class="color3">$\lvert f(x)-\ell \rvert < \varepsilon$</span>.

#### Or, in words,
            <span class="color1">No matter how small we pick $\varepsilon>0$</span>,

<span class="color6">we can always go close enough to $a$ (ie. $\delta$ close) to ensure that</span>

<span class="color5">all $x$ closer to $a$ (ie. within $\delta$ of $a$) give</span>
            <span class="color3">$f(x)$ within $\varepsilon$ of $\ell$</span>.
            #### How to prove convergence if you know the limit
            Construct a machine (a function) that <span class="color6">tells you $\delta$</span> <span class="color1">for any given $\varepsilon$</span>, and that $\delta$ <span class="color3">d</span><span class="color5">o</span><span class="color3">e</span><span class="color5">s</span><span class="color3"> t</span><span class="color5">h</span><span class="color3">e</span><span class="color5"> j</span><span class="color3">o</span><span class="color5">b</span>.
            
            ---
            ## Limits of real functions
            #### Definition
            $f(x)$ *converges* to *limit* $\ell\in\mathbb R$ as $x\to a$ if

#### How to prove convergence if you know the limit
            Construct a machine (a function) that <span class="color6">tells you $\delta$</span> <span class="color1">for any given $\varepsilon$</span>, and that $\delta$ <span class="color3">d</span><span class="color5">o</span><span class="color3">e</span><span class="color5">s</span><span class="color3"> t</span><span class="color5">h</span><span class="color3">e</span><span class="color5"> j</span><span class="color3">o</span><span class="color5">b</span>.

#### Theorem
            If $f(x)=x^2$ then $\lim_{x\to0}f(x) = 0$.
            #### Proof
            For any $\varepsilon>0$, let $\delta=\sqrt\varepsilon$.
            If $\lvert x-0 \rvert \lt \delta$, then
            $\left\lvert f(x)-0 \right\rvert = \lvert x^2 \rvert = x^2 \lt \delta^2 = \varepsilon$.   □
            
            ---
            ## Limits of real functions
            #### Definition
            $f(x)$ *converges* to *limit* $\ell\in\mathbb R$ as $x\to a$ if

#### Theorem
            If $f(x)=x^2$ then $\lim_{x\to0}f(x) = 0$.
            #### Rough work
            I want to make
            $\left\lvert f(x)-0 \right\rvert \lt \varepsilon$.
            I calculate $\left\lvert f(x)-0 \right\rvert = \left\lvert x^2 \right\rvert = x^2$

and I'm allowed to insist $\lvert x-0 \rvert \lt \delta$, which implies $x^2 \lt \delta^2$.

So I just need $\delta^2 \leqslant \varepsilon$.
            So choose $\delta = \sqrt\varepsilon$.
            
            ---
            ## Limits of real functions
            #### Definition
            $f(x)$ *converges* to *limit* $\ell\in\mathbb R$ as $x\to a$ if

and I'm allowed to insist $\lvert x-0 \rvert \lt \delta$, which implies $x^2 \lt \delta^2$.

So I just need $\delta^2 \leqslant \varepsilon$.
            So choose $\delta = \sqrt\varepsilon$.
            #### Proof
            For any $\varepsilon>0$, let $\delta=\sqrt\varepsilon$.
            If $\lvert x-0 \rvert \lt \delta$, then
            $\left\lvert f(x)-0 \right\rvert = \lvert x^2 \rvert = x^2 \lt \delta^2 = \varepsilon$.   □
            
            ---
            ## Limits of real functions
            #### Definition
            $f(x)$ *converges* to *limit* $\ell\in\mathbb R$ as $x\to a$ if

#### Theorem
            If $f(x)=\frac{x+1}{2-x}$ then $\lim_{x\to1}f(x) = 2$.
            #### Proof
            For any $\varepsilon>0$, let $\delta=\min\\{\frac\varepsilon6,\frac12\\}$.

If $\lvert x-1 \rvert \lt \delta$, then $6\lvert x-1 \rvert\lt\varepsilon$ and $\lvert 2-x \rvert = 2-x \gt \frac12$,

so
            $\left\lvert\frac{x+1}{2-x} - 2 \right\rvert = \frac{3\lvert x-1 \rvert}{\lvert 2-x \rvert} \lt \frac{3\lvert x-1 \rvert}{\frac12} = 6 \lvert x-1 \rvert\lt\varepsilon$.   □
            
            ---
            ## Limits of real functions
            #### Theorem
            If $f(x)=\frac{x+1}{2-x}$ then $\lim_{x\to1}f(x) = 2$.
            #### Rough work
            I want to make
            $\left\lvert f(x)-2 \right\rvert \lt \varepsilon$.
            I calculate $\left\lvert f(x)-2 \right\rvert = \ldots = \frac{3\lvert x-1 \rvert}{\lvert 2-x \rvert}$

and I'm allowed to insist $\lvert x-1 \rvert \lt \delta$.
            If $x$ is close to $1$ then $\lvert 2-x \rvert$ is close to $1$ too.

Quantitatively, if $\lvert x-1 \rvert \lt \frac12$, then $\frac12 \lt \lvert 2-x \rvert \lt \frac32$, so $\frac1{\lvert 2-x \rvert} \lt 2$.

That tells me $\left\lvert f(x)-2 \right\rvert = \frac{3\lvert x-1 \rvert}{\lvert 2-x \rvert} \lt 6\lvert x-1 \rvert$.

Now to achieve $\left\lvert f(x)-2 \right\rvert \lt \varepsilon$, all I need is $6\lvert x-1 \rvert \lt \varepsilon$; equivalently $\lvert x-1 \rvert \lt \frac\varepsilon6$.

To make all this work, I need both $\lvert x-1 \rvert \lt \frac12$ and $\lvert x-1 \rvert \lt \frac\varepsilon6$.

So choose $\delta = \min\\{\frac\varepsilon6,\frac12\\}$.
            
            ---
            ## Limits of real functions
            ### Why not input at the limit point?
            #### Definition
            $f(x)$ *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$.

Why $0 \lt \lvert x-a \rvert \lt \delta$ not just $\lvert x-a \rvert \lt \delta$?

Consider $\chi\_{\\{3\\}}:\mathbb R\to\mathbb R$, the indicator function of the set $\\{3\\}$, which is $0$ everywhere except $1$ if the input is $3$.
            </div>
            <div style="flex:1;margin-left:2vw">

![Desmos plot of y=indicator function for set {3} of x](gfx/chi3.png "Desmos plot of y=\chi_{3}(x)")
            </div>
            </div>

We want limits to be defined so that $\displaystyle \lim\_{x\to3}\chi\_{\\{3\\}}(x) = 0$.
            
            ---
            ## Limits of real functions
            #### Definition
            $f(x)$ *converges* to *limit* $\ell\in\mathbb R$ as $x\to a$ if

#### Theorem
            If $f(x)=5x+3-\chi\_{\\{2\\}}(x)$ then $f(2)=12$ but $\lim_{x\to2}f(x) = ?$.
            #### Proof
            $f(2) = 5(2)+3-\chi\_{\\{2\\}}(1) = 10+3-1=12$.

Your turn!
            $\ldots$
            #### How to prove convergence if you know the limit
            Construct a machine (a function) that <span class="color6">tells you $\delta$</span> <span class="color1">for any given $\varepsilon$</span>, and that $\delta$ <span class="color3">d</span><span class="color5">o</span><span class="color3">e</span><span class="color5">s</span><span class="color3"> t</span><span class="color5">h</span><span class="color3">e</span><span class="color5"> j</span><span class="color3">o</span><span class="color5">b</span>.
            
            ---
            ## Limits of real functions
            #### Theorem
            If $f(x)=5x+3-\chi\_{\\{2\\}}(x)$ then $f(2)=12$ but $\lim_{x\to2}f(x) = ?$.
            #### Rough work
            If $x\approx2$ but $x\neq2$, then $f(x)\approx 5 (2) + 3 - 0 = 13$, so guess that the limit is $13$.

I want to make
            $\left\lvert f(x)-13 \right\rvert \lt \varepsilon$.
            Suppose $x\neq2$.
            Then $f(x)=5x+3-0$.

I calculate $\left\lvert f(x)-13 \right\rvert = \left\lvert 5x+3-13 \right\rvert = 5\left\lvert x-2 \right\rvert$

and I'm allowed to insist $0 \lt \lvert x-2 \rvert \lt \delta$.
            Now to achieve $\left\lvert f(x)-13 \right\rvert \lt \varepsilon$, all I need is $5\lvert x-2 \rvert \lt \varepsilon$; equivalently $\lvert x-2 \rvert \lt \frac\varepsilon5$.
            So choose $\delta = \frac\varepsilon5$.
            
            ---
            ## Limits of real functions
            #### Theorem
            If $f(x)=5x+3-\chi\_{\\{2\\}}(x)$ then $f(2)=12$ but $\lim_{x\to2}f(x) = ?$.
            #### Rough work
            If $x\approx2$ but $x\neq2$, then $f(x)\approx 5 (2) + 3 - 0 = 13$, so guess that the limit is $13$.

I want to make
            $\left\lvert f(x)-13 \right\rvert \lt \varepsilon$.
            Suppose $x\neq2$.
            Then $f(x)=5x+3-0$.

I calculate $\left\lvert f(x)-13 \right\rvert = \left\lvert 5x+3-13 \right\rvert = 5\left\lvert x-2 \right\rvert$

and I'm allowed to insist $0 \lt \lvert x-2 \rvert \lt \delta$.
            Now to achieve $\left\lvert f(x)-13 \right\rvert \lt \varepsilon$, all I need is $5\lvert x-2 \rvert \lt \varepsilon$; equivalently $\lvert x-2 \rvert \lt \frac\varepsilon5$.
            So choose $\delta = \frac\varepsilon5$.
            #### Proof
            For any $\varepsilon>0$, let $\delta=\frac\varepsilon5$.
            If $0 \lt \lvert x-2 \rvert \lt \delta$, then $\left\lvert f(x)-13 \right\rvert = \left\lvert 5x+3-\chi\_{\\{2\\}}(x)-13 \right\rvert = 5 \lvert x-2 \rvert\lt5\frac\varepsilon5=\varepsilon$.   □
            
            ---
            ## Limits of real functions
            <div style="display:flex;flex-direction:row;width:100%">
            <div style="flex:1 1 0px;margin-right:2vw;background-color:white;color:black">

Plot of $y=f(x)$.

![Desmos plot of y=x plus sgn of x minus 2](gfx/xplussgnxminus2.png "Desmos plot of y=x+sgn(x-2)")

</div>
            <div style="flex:0.3 0.3 0px;margin-left:2vw">

$\displaystyle f(2)=$

$\displaystyle\lim\_{x\to3}f(x)=$

$\displaystyle\lim\_{x\to2}f(x)=$ 
            </div>
            <div style="flex:1.7 1.7 0px;margin-left:2vw">

</div>
            </div>
            
            ---
            ## Limits of real functions
            <div style="display:flex;flex-direction:row;width:100%">
            <div style="flex:1 1 0px;margin-right:2vw;background-color:white;color:black">

Plot of $y=f(x)$.

![Desmos plot of y=x plus sgn of x minus 2](gfx/xplussgnxminus2.png "Desmos plot of y=x+sgn(x-2)")

</div>
            <div style="flex:0.3 0.3 0px;margin-left:2vw">

$\displaystyle f(2)=$

$\displaystyle\lim\_{x\to3}f(x)=$

$\displaystyle\lim\_{x\to2}f(x)=$ 
            </div>
            <div style="flex:1.7 1.7 0px;margin-left:2vw">

$\displaystyle 2 \phantom{f(2)}$

$\displaystyle 4 \phantom{\lim\_{x\to3}}$

undefined $\displaystyle \phantom{\lim\_{x\to2}}$
            </div>
            </div>
            
            ---
            ## Limits of real functions
            <div style="display:flex;flex-direction:row;width:100%">
            <div style="flex:1 1 0px;margin-right:2vw;background-color:white;color:black">

Plot of $y=f(x)$.

![Desmos plot of y=x plus sgn of x minus 2](gfx/xplussgnxminus2.png "Desmos plot of y=x+sgn(x-2)")

</div>
            <div style="flex:0.3 0.3 0px;margin-left:2vw">

$\displaystyle f(2)=$

$\displaystyle\lim\_{x\to3}f(x)=$

$\displaystyle\lim\_{x\to2}f(x)=$ 
            </div>
            <div style="flex:1.7 1.7 0px;margin-left:2vw">

$\displaystyle 2 \phantom{f(2)}$

$\displaystyle 4 \phantom{\lim\_{x\to3}}$

undefined $\displaystyle \phantom{\lim\_{x\to2}}$

but it "should" be $1$ or $3$, depending on which side we approach from.
            </div>
            </div>
            
            ---
            ## Limits of real functions
            <div style="display:flex;flex-direction:row;width:100%">
            <div style="flex:1 1 0px;margin-right:2vw;background-color:white;color:black">

Plot of $y=f(x)$.

![Desmos plot of y=x plus sgn of x minus 2](gfx/xplussgnxminus2.png "Desmos plot of y=x+sgn(x-2)")

</div>
            <div style="flex:1.99 1.99 0px;margin-left:2vw">

#### Notation
              One sided limits are:

*the limit from the left* $\displaystyle \lim\_{x\to a^-} f(x)$ and

*the limit from the right* $\displaystyle \lim\_{x\to a^+} f(x)$.

#### Example
              for $f$ on the left,
              $\displaystyle \lim\_{x\to 2^-} f(x) = 1$
              and
              $\displaystyle \lim\_{x\to 2^+} f(x) = 3$.

#### Theorem
              $\displaystyle \lim\_{x\to a} f(x)$ exists if and only if both one sided limits exist and $\displaystyle \lim\_{x\to a^-} f(x) = \lim\_{x\to a^+} f(x)$.
            </div>
            <div style="flex:0.01 0.01 0px;margin-left:2vw">

</div>
            </div>
            
            ---
            ## Definitions of one sided limits
            #### Definition (recall; two sided limit)
            $f(x)$ *converges* to *limit* $\ell\in\mathbb R$ as $x\to a$ if

<span class="color1">for all $\varepsilon>0$</span>,   <span class="color6">$\exists \delta>0$ such that</span>   <span class="color5 bg-highlight">$0 \lt \lvert x-a \rvert \lt \delta \implies$</span> <span class="color3">$\lvert f(x)-\ell \rvert < \varepsilon$</span>.
            #### Definition (limit from the left)
            $f(x)$ *converges* to *limit from the left* $\ell\in\mathbb R$ as $x\to a^-$ if

<span class="color1">for all $\varepsilon>0$</span>,   <span class="color6">$\exists \delta>0$ such that</span>   <span class="color5 bg-highlight">$0 \lt a-x \lt \delta \implies$</span> <span class="color3">$\lvert f(x)-\ell \rvert < \varepsilon$</span>.
            #### Definition (limit from the right)
            $f(x)$ *converges* to *limit from the right* $\ell\in\mathbb R$ as $x\to a^+$ if

<span class="color1">for all $\varepsilon>0$</span>,   <span class="color6">$\exists \delta>0$ such that</span>   <span class="color5 bg-highlight">$0 \lt x-a \lt \delta \implies$</span> <span class="color3">$\lvert f(x)-\ell \rvert \lt \varepsilon$</span>.
            
            ---
            ## Algebra of limits
            If $\displaystyle\lim\_{x\to a}f(x)$ and $\displaystyle\lim\_{x\to a}g(x)$ both exist and $c\in\mathbb R$, then
            1. $\displaystyle\lim\_{x\to a}(f(x) \pm g(x)) = \left(\displaystyle\lim\_{x\to a}f(x)\right) \pm \left(\displaystyle\lim\_{x\to a}g(x)\right)$.<br><br>
            1. $\displaystyle\lim\_{x\to a}(f(x) g(x)) = \left(\displaystyle\lim\_{x\to a}f(x)\right) \left(\displaystyle\lim\_{x\to a}g(x)\right)$.<br><br>
            1. $\displaystyle\lim\_{x\to a}\left(\frac{f(x)}{g(x)}\right) = \frac{\displaystyle\lim\_{x\to a}f(x)}{\displaystyle\lim\_{x\to a}g(x)}$, provided $\displaystyle\lim\_{x\to a}g(x)\neq0$.<br><br>
            1. $\displaystyle\lim\_{x\to a}cf(x)) = c\left(\displaystyle\lim\_{x\to a}f(x)\right)$.

In each of these, $x\to a$ can be replaced by $x\to a^\pm$.
            
            ---
            ## Squeeze theorem
            Suppose $f,g,h$ are functions defined on an open interval $I$, except possibly at $a\in I$, and, for all $x\in I$, $f(x) \leqslant g(x) \leqslant h(x)$.

If $\displaystyle\lim\_{x\to a}f(x) = \ell = \displaystyle\lim\_{x\to a}h(x)$,

then also $\displaystyle\lim\_{x\to a}g(x)=\ell$.

There are also versions for one sided limits.
            
            ---
            ## Example
            <div style="display:flex;flex-direction:row;width:100%">
            <div style="flex:0.8 0.8 0px;margin-right:2vw;background-color:white;color:black">

Plot $y=f(x)$

![Desmos plot of y=x plus sgn of x minus 2 plus 2x minus 4 times sin of 5 on x minus 2](gfx/xplussgnxminus2plus2xminus4sin5onxminus2.png "Desmos plot of y=x+sgn(x-2)+2(x-2)sin(5/(x-2))")

</div>
            <div style="flex:0.15 0.15 0px;margin-left:2vw">

$\displaystyle \phantom{\left(\frac52\right)}f(x) =$

$\displaystyle\lim\_{x\to2^-}f(x)=$

$\displaystyle\lim\_{x\to2^+}f(x)=$

$\displaystyle\lim\_{x\to2}f(x)=$ 
            </div>
            <div style="flex:2 2 0px;margin-left:2vw">

$\displaystyle \operatorname{sgn}(x-2) + x + 2(x-2)\sin\left(\frac5{x-2}\right)$

$\displaystyle ? \phantom{\lim\_{2^-}}$

$\displaystyle ? \phantom{\lim\_{2^+}}$

$\displaystyle ? \phantom{\lim\_{2}}$
            </div>
            </div>
            
            ---
            ## Example
            <div style="display:flex;flex-direction:row;width:100%">
            <div style="flex:0.8 0.8 0px;margin-right:2vw;background-color:white;color:black">

Plot $y=f(x)$

![Desmos plot of y=x plus sgn of x minus 2 plus 2x minus 4 times sin of 5 on x minus 2](gfx/xplussgnxminus2plus2xminus4sin5onxminus2.png "Desmos plot of y=x+sgn(x-2)+2(x-2)sin(5/(x-2))")

</div>
            <div style="flex:0.15 0.15 0px;margin-left:2vw">

$\displaystyle \phantom{\left(\frac52\right)}f(x) =$

$\displaystyle\lim\_{x\to2^-}f(x)=$

$\displaystyle\lim\_{x\to2^+}f(x)=$

$\displaystyle\lim\_{x\to2}f(x)=$ 
            </div>
            <div style="flex:2 2 0px;margin-left:2vw">

$\displaystyle \operatorname{sgn}(x-2) + x + 2(x-2)\sin\left(\frac5{x-2}\right)$

$\displaystyle 1 \phantom{\lim\_{2^-}}$

$\displaystyle 3 \phantom{\lim\_{2^+}}$

undefined $\displaystyle \phantom{\lim\_{2}}$
            </div>
            </div>

---
            ## Example
            Let $\displaystyle f(x)= \operatorname{sgn}(x-2) + x + 2(x-2)\sin\left(\frac5{x-2}\right)$.
            Then $\displaystyle\lim\_{x\to2^+}f(x)=3$.
            #### Proof (direct from definition)
            For any $\varepsilon\gt0$, let $\delta=\frac\varepsilon3$, and suppose $0 \lt x-2 \lt \delta$.

Then $x \gt 2$, so $\operatorname{sgn}(x-2)=1$, so
            \\[\begin{aligned}
              \left\lvert f(x) - 3 \right\rvert
              &= \left\lvert 1+x+2(x-2)\sin\left(\frac5{x-2}\right) - 3 \right\rvert \\\\
              &= \left\lvert x+2(x-2)\sin\left(\frac5{x-2}\right) - 2 \right\rvert \\\\
              &= (x-2) \left\lvert 1 + 2\sin\left(\frac5{x-2}\right) \right\rvert
              \leqslant (x-2) 3 \\\\
              &\lt 3 \delta = 3\frac\varepsilon3=\varepsilon.
              \qquad □
            \end{aligned}\\]
            
            ---
            ## Example
            Let $\displaystyle f(x)= \operatorname{sgn}(x-2) + x + 2(x-2)\sin\left(\frac5{x-2}\right)$.
            Then $\displaystyle\lim\_{x\to2^+}f(x)=3$.
            #### Proof (attempt using limit laws)
            $\displaystyle\lim\_{x\to2^+}f(x) = \lim\_{x\to2^+}\left[ \operatorname{sgn}(x-2)+x+2(x-2)\sin\left(\frac5{x-2}\right) \right]$

$\displaystyle= \lim\_{x\to2^+}[\operatorname{sgn}(x-2)] + \lim\_{x\to2^+}[x] + 2 \times \lim\_{x\to2^+}\left[ x-2 \right] \times \lim\_{x\to2^+}\left[ \sin\left(\frac5{x-2}\right) \right]$

$\displaystyle= 1 + 2 + 2\times 0 \times \lim\_{x\to2^+}\left[ \sin\left(\frac5{x-2}\right) \right]$,
            provided the last limit exists.
            
            ---
            ## Example
            Plot of $\displaystyle y=\sin\left(\frac5{x-2}\right)$.

![Desmos plot of y=sin of 5 on x minus 2](gfx/sin5onxminus2.png "Desmos plot of y=sin(5/(x-2))")

So this limit does not exist.
            
            ---
            ## Example
            Let $\displaystyle f(x)= \operatorname{sgn}(x-2) + x + 2(x-2)\sin\left(\frac5{x-2}\right)$.
            Then $\displaystyle\lim\_{x\to2^+}f(x)=3$.
            #### Proof (attempt using limit laws)
            $\displaystyle\lim\_{x\to2^+}f(x) = \lim\_{x\to2^+}\left[ \operatorname{sgn}(x-2)+x+2(x-2)\sin\left(\frac5{x-2}\right) \right]$

$\displaystyle= \lim\_{x\to2^+}[\operatorname{sgn}(x-2)] + \lim\_{x\to2^+}[x] + 2 \times \lim\_{x\to2^+}\left[ x-2 \right] \times \lim\_{x\to2^+}\left[ \sin\left(\frac5{x-2}\right) \right]$

$\displaystyle= 1 + 2 + 2\times 0 \times \lim\_{x\to2^+}\left[ \sin\left(\frac5{x-2}\right) \right]$,
            but this did not work.
            
            ---
            ## Example
            Let $\displaystyle f(x)= \operatorname{sgn}(x-2) + x + 2(x-2)\sin\left(\frac5{x-2}\right)$.
            Then $\displaystyle\lim\_{x\to2^+}f(x)=3$.
            #### Proof (using limit laws and squeeze theorem)
            $\displaystyle\lim\_{x\to2^+}f(x) = \lim\_{x\to2^+}\left[ \operatorname{sgn}(x-2)+x+2(x-2)\sin\left(\frac5{x-2}\right) \right]$

$\displaystyle= \lim\_{x\to2^+}[\operatorname{sgn}(x-2)] + \lim\_{x\to2^+}[x] + 2 \times \lim\_{x\to2^+}\left[ (x-2)\sin\left(\frac5{x-2}\right) \right]$

$\displaystyle= 1 + 2 + 2 \times \lim\_{x\to2^+}\left[ (x-2)\sin\left(\frac5{x-2}\right) \right]$,
            provided the last limit exists.
            
            ---
            ## Example
            Let $\displaystyle f(x)= \operatorname{sgn}(x-2) + x + 2(x-2)\sin\left(\frac5{x-2}\right)$.
            Then $\displaystyle\lim\_{x\to2^+}f(x)=3$.
            #### Proof (using limit laws and squeeze theorem)
            $\displaystyle\lim\_{x\to2^+}f(x) = \lim\_{x\to2^+}\left[ \operatorname{sgn}(x-2)+x+2(x-2)\sin\left(\frac5{x-2}\right) \right]$

$\displaystyle= \lim\_{x\to2^+}[\operatorname{sgn}(x-2)] + \lim\_{x\to2^+}[x] + 2 \times \lim\_{x\to2^+}\left[ (x-2)\sin\left(\frac5{x-2}\right) \right]$

$\displaystyle= 1 + 2 + 2 \times \lim\_{x\to2^+}\left[ (x-2)\sin\left(\frac5{x-2}\right) \right]$,
            provided the last limit exists.

But
            $-1 \lt \sin\left(\frac5{x-2}\right) \lt 1$, so $-(x-2) \lt (x-2)\sin\left(\frac5{x-2}\right) \lt (x-2)$,
            and $\lim\_{x\to2^+} [-(x-2)] = 0 = \lim\_{x\to2^+} (x-2)$,
            so the last limit does exist, and is zero.

Hence $\displaystyle\lim\_{x\to2^+}f(x) = 1+2 + 2\times0 = 3$.    □      What needs changing for $x\to2^-$?