Show $|e^{2x\pi i}-1|\leq 2\pi |x|$ for real $x$.
We first prove another result: $$\bigg| \int_b^a f(t) dt\bigg|\leq \int_b^a |f(t)| dt.$$ $$\bigg| \int_b^a f(t) dt\bigg|\stackrel{(*)}{=}e^{-i\phi}\int_b^a f(t) dt=\int_b^a e^{-i\phi}f(t) dt$$ $(*)\quad F=|F|e^{i\phi}$ and $e^{i\phi}e^{-i\phi}=1$ means $|F|=Fe^{-i\phi}$.
Since $\bigg| \int_b^a f(t) dt\bigg|$ is real and $\text{Re}(z)\leq |z|$, we have $$\begin{align}\bigg| \int_b^a f(t) dt\bigg|&=\text{Re}\bigg(\int_b^a e^{-i\phi}f(t) dt\bigg)\\&=\int_b^a \text{Re}(e^{-i\phi}f(t)) dt\\&\leq \int_b^a |e^{-i\phi}f(t)| dt\\&=\int_b^a |f(t)| dt.\end{align}$$ One can also prove this using Cauchy Schwarz inequality.
Onto the problem, let $f(t)=e^{ixt}$ for $x,t\in \Bbb R$. $$\bigg|\int_0^{2\pi}e^{ixt}dt\bigg|\leq \int_0^{2\pi}|e^{ixt}|dt=2\pi$$ Also $$\bigg|\int_0^{2\pi}e^{ixt}dt\bigg|=\bigg[\bigg|\frac{e^{ixt}}{ix}\bigg|\bigg]_0^{2\pi}=\frac{|e^{2x\pi i}-1|}{|x|}.$$ Therefore $$|e^{2x\pi i}-1|\leq 2\pi |x|.$$
Saturday, 24 October 2015
Sunday, 11 October 2015
A problem in complex analysis
Show that if $f$ is holomorphic, then $$\Delta|f(z)+z|^2=4|f'(z)+1|^2.$$
There are two ways to deal with the modulus: either use the formula $|z|^2=z\bar{z}$ or separate $z$ into real and imaginary parts. For ease of calculations, we separate $f(z)+z$ into real and imaginary parts, having $$\Delta|f(z)+z|^2=\Delta\bigg((u+x)^2+(v+y)^2\bigg).$$ [Here $u,v$ are actually $u(x,y)$ and $v(x,y)$ respectively.] $$\partial_x \bigg((u+x)^2+(v+y)^2\bigg)=2(u+x)(u_x+1)+2(v+y)v_x\\ \partial_x^2 \bigg((u+x)^2+(v+y)^2\bigg)=2(u+x)u_{xx}+2(v+y)v_{xx}+2(u_x+1)^2+2v_x^2$$ Therefore, $$\Delta\bigg((u+x)^2+(v+y)^2\bigg)=2(u+x)(u_{xx}+u_{yy})+2(v+y)(v_{xx}+v_{yy})\\+2\bigg( (u_x+1)^2+v_x^2+u_y^2+(v_y+1)^2\bigg).$$ Since $f$ is holomorphic, we have $$u_{xx}+u_{yy}=0\\ v_{xx}+v_{yy}=0\\ u_x=v_y,\quad u_y=-v_x.$$ Finally, $$\Delta|f(z)+z|^2=4\bigg( (u_x+1)^2+u_y^2\bigg)=4|f'(z)+1|^2$$ since $f'(z)+1=u_x+1+iu_y$ and $|f'(z)+1|=(u_x+1)^2+u_y^2$.
There are two ways to deal with the modulus: either use the formula $|z|^2=z\bar{z}$ or separate $z$ into real and imaginary parts. For ease of calculations, we separate $f(z)+z$ into real and imaginary parts, having $$\Delta|f(z)+z|^2=\Delta\bigg((u+x)^2+(v+y)^2\bigg).$$ [Here $u,v$ are actually $u(x,y)$ and $v(x,y)$ respectively.] $$\partial_x \bigg((u+x)^2+(v+y)^2\bigg)=2(u+x)(u_x+1)+2(v+y)v_x\\ \partial_x^2 \bigg((u+x)^2+(v+y)^2\bigg)=2(u+x)u_{xx}+2(v+y)v_{xx}+2(u_x+1)^2+2v_x^2$$ Therefore, $$\Delta\bigg((u+x)^2+(v+y)^2\bigg)=2(u+x)(u_{xx}+u_{yy})+2(v+y)(v_{xx}+v_{yy})\\+2\bigg( (u_x+1)^2+v_x^2+u_y^2+(v_y+1)^2\bigg).$$ Since $f$ is holomorphic, we have $$u_{xx}+u_{yy}=0\\ v_{xx}+v_{yy}=0\\ u_x=v_y,\quad u_y=-v_x.$$ Finally, $$\Delta|f(z)+z|^2=4\bigg( (u_x+1)^2+u_y^2\bigg)=4|f'(z)+1|^2$$ since $f'(z)+1=u_x+1+iu_y$ and $|f'(z)+1|=(u_x+1)^2+u_y^2$.
Thursday, 1 October 2015
Harmonic conjugate and analyticity
a) For which values of $a,b,c,d$ is the function $u(x,y)=ax^3+bx^2y+cxy^2+dy^3$ harmonic?
b) Find a harmonic conjugate $v(x,y)$ of $u(x,y)$ in this case.
c) Find an analytic function $f(z)=u(x,y)+iv(x,y)$, where $z=x+iy$.
a) Use the Laplace equation for $u$ to get $$(\partial_x^2+\partial_y^2)u(x,y)=(6a+2c)x+(2b+6d)y=0,$$ thus $c=-3a,b=-3d$.
b) $$u(x,y)=ax^3-3dx^2y-3axy^2+dy^3.$$ Harmonic conjugate satisfies the Cauchy-Riemann condition $$v_y=u_x=3a(x^2-y^2)-6dxy,\\ v_x=-u_y=3d(x^2-y^2)+6axy.$$ Integrating these, we have $$v(x,y)=dx^3+3ax^2y-3dxy^2-ay^3.$$ Alternatively, we can put $$f(z)=2u(\dfrac{z}{2},\dfrac{z}{2i})-u(0,0).$$ Then $$\begin{align}f(z)&=2\bigg(a(\dfrac{z}{2})^3-3d(\dfrac{z}{2})^2(\dfrac{z}{2i})-3a(\dfrac{z}{2})(\dfrac{z}{2i})^2+d(\dfrac{z}{2i})^3\bigg)\\&=(a+3id+3a+id)\dfrac{z^3}{4}\\&=(a+id)z^3\\&=(a+id)(x+iy)^3\end{align}$$ The imaginary part is $$v(x,y)=dx^3+3ax^2y-3dxy^2-ay^3.$$ c) $$\begin{align}f(z)&=u(x,y)+iv(x,y)\\&=ax^3-3dx^2y-3axy^2+dy^3+i(dx^3+3ax^2y-3dxy^2-ay^3)\\&=a(x^3+3ix^2y-3xy^2-iy^3)+d(ix^3-3x^2y-3ixy^2+y^3)\\&=a(x+iy)^3+id(x+iy)^3\\&=(a+id)z^3\end{align}$$ Remark: The coefficients $1,3,3,1$ hint the use of $(?+?)^3$. We can also substitute $y=0,x=z$ to get $(a+id)z^3$ immediately!
b) Find a harmonic conjugate $v(x,y)$ of $u(x,y)$ in this case.
c) Find an analytic function $f(z)=u(x,y)+iv(x,y)$, where $z=x+iy$.
a) Use the Laplace equation for $u$ to get $$(\partial_x^2+\partial_y^2)u(x,y)=(6a+2c)x+(2b+6d)y=0,$$ thus $c=-3a,b=-3d$.
b) $$u(x,y)=ax^3-3dx^2y-3axy^2+dy^3.$$ Harmonic conjugate satisfies the Cauchy-Riemann condition $$v_y=u_x=3a(x^2-y^2)-6dxy,\\ v_x=-u_y=3d(x^2-y^2)+6axy.$$ Integrating these, we have $$v(x,y)=dx^3+3ax^2y-3dxy^2-ay^3.$$ Alternatively, we can put $$f(z)=2u(\dfrac{z}{2},\dfrac{z}{2i})-u(0,0).$$ Then $$\begin{align}f(z)&=2\bigg(a(\dfrac{z}{2})^3-3d(\dfrac{z}{2})^2(\dfrac{z}{2i})-3a(\dfrac{z}{2})(\dfrac{z}{2i})^2+d(\dfrac{z}{2i})^3\bigg)\\&=(a+3id+3a+id)\dfrac{z^3}{4}\\&=(a+id)z^3\\&=(a+id)(x+iy)^3\end{align}$$ The imaginary part is $$v(x,y)=dx^3+3ax^2y-3dxy^2-ay^3.$$ c) $$\begin{align}f(z)&=u(x,y)+iv(x,y)\\&=ax^3-3dx^2y-3axy^2+dy^3+i(dx^3+3ax^2y-3dxy^2-ay^3)\\&=a(x^3+3ix^2y-3xy^2-iy^3)+d(ix^3-3x^2y-3ixy^2+y^3)\\&=a(x+iy)^3+id(x+iy)^3\\&=(a+id)z^3\end{align}$$ Remark: The coefficients $1,3,3,1$ hint the use of $(?+?)^3$. We can also substitute $y=0,x=z$ to get $(a+id)z^3$ immediately!
Tuesday, 22 September 2015
$|z-1|\leq \bigg| |z|-1\bigg|+|z| |Arg\:z|$
For a non-zero $z$ and $-\pi<Arg\:z\leq \pi$, show that $$|z-1|\leq \bigg| |z|-1\bigg|+|z| |Arg\:z|.$$
$$\begin{align}|z-1|&=\bigg|z-|z|+|z|-1\bigg|\\
&\leq \bigg|z-|z|\bigg|+\bigg||z|-1\bigg|\\
&=|z||\cos\theta+i\sin\theta-1|+\bigg||z|-1\bigg|\\
&=|z|\sqrt{(\cos\theta-1)^2+\sin^2\theta}+\bigg||z|-1\bigg|\\
&=|z|\left|2\sin\frac{\theta}{2}\right|+\bigg||z|-1\bigg|\\
&\leq |z||Arg\:z|+\bigg||z|-1\bigg|\end{align}$$ since $(\cos\theta-1)^2+\sin^2\theta=(1-2\sin^2\dfrac{\theta}{2}-1)^2+4\sin^2\dfrac{\theta}{2}\cos^2\dfrac{\theta}{2}=4\sin^2\dfrac{\theta}{2}$ and $\bigg|\sin\dfrac{\theta}{2}\bigg|<\bigg|\dfrac{\theta}{2}\bigg|$.
Geometric interpretation
Take $|z|>1$. Consider the triangle with vertices $1,z,|z|$. By the triangle inequality, we have $|z-1|\leq \bigg|z-|z|\bigg|+\bigg||z|-1\bigg|$. Recall the arc length formula $s=r\theta$. Here, $|z| |Arg\:z|$ is actually the arc length joining $z$ and $|z|$, which is larger than the chord joining the two points. Therefore, the inequality holds.
$$\begin{align}|z-1|&=\bigg|z-|z|+|z|-1\bigg|\\
&\leq \bigg|z-|z|\bigg|+\bigg||z|-1\bigg|\\
&=|z||\cos\theta+i\sin\theta-1|+\bigg||z|-1\bigg|\\
&=|z|\sqrt{(\cos\theta-1)^2+\sin^2\theta}+\bigg||z|-1\bigg|\\
&=|z|\left|2\sin\frac{\theta}{2}\right|+\bigg||z|-1\bigg|\\
&\leq |z||Arg\:z|+\bigg||z|-1\bigg|\end{align}$$ since $(\cos\theta-1)^2+\sin^2\theta=(1-2\sin^2\dfrac{\theta}{2}-1)^2+4\sin^2\dfrac{\theta}{2}\cos^2\dfrac{\theta}{2}=4\sin^2\dfrac{\theta}{2}$ and $\bigg|\sin\dfrac{\theta}{2}\bigg|<\bigg|\dfrac{\theta}{2}\bigg|$.
Geometric interpretation
Take $|z|>1$. Consider the triangle with vertices $1,z,|z|$. By the triangle inequality, we have $|z-1|\leq \bigg|z-|z|\bigg|+\bigg||z|-1\bigg|$. Recall the arc length formula $s=r\theta$. Here, $|z| |Arg\:z|$ is actually the arc length joining $z$ and $|z|$, which is larger than the chord joining the two points. Therefore, the inequality holds.
Saturday, 19 September 2015
Discovery work in analysis
In this post, we will deal with statements of the form $$\forall x\;\exists y \bullet P(x,y).$$ To prove these statements, we need some preliminary 'discovery work' before embarking on the proof.
Convergence of sequences
Definition: The sequence $(a_n)$ converges to $l$ if, for any $\epsilon>0$, there exists $N\in \Bbb Z^+$ such that $$n>N \Rightarrow |a_n-l|<\epsilon.$$ We can write either $a_n\to l$ as $n\to \infty$ or $\lim\limits_{n\to \infty}a_n=l$.
Example
A sequence $(a_n)$ is defined by $a_n=\dfrac{3n^2-4n}{(n+1)(n+2)}$ for $n\in \Bbb Z^+$. Show that $\lim\limits_{n\to \infty}a_n=3$.
The definition of convergence involves two quantifiers and is of the form $$\forall \epsilon\;\exists N \bullet P(\epsilon,N)$$ where $P(\epsilon,N)$ is a propositional function and the universes for $\epsilon$ and $N$ are $\Bbb R^+$ and $\Bbb Z^+$ respectively. The structure of the proof starts with an arbitrary $\epsilon$ (in its universe) and then selects a particular $N$ (in its universe) which may depend on $\epsilon$. To complete the proof, we need to show that the propositional function $P(\epsilon,N)$ is satisfied for the arbitrary $\epsilon$ and particular $N$.
We first start by the definition. We need to find a positive integer $N$ such that for all $n>N$, $$|a_n-l|=\left|\frac{3n^2-4n}{(n+1)(n+2)}-3 \right|<\epsilon.$$ But how? We can manipulate $|a_n-l|$ until it is less than any positive $\epsilon$: $$\begin{align}\left| \dfrac{3n^2-4n}{(n+1)(n+2)}-3 \right|&=\left| \dfrac{3n^2-4n-3(n+1)(n+2)}{(n+1)(n+2)} \right|\\&=\left| \dfrac{3n^2-4n-(3n^2+9n+6)}{(n+1)(n+2)} \right|\\&=\left| \dfrac{-13n-6}{(n+1)(n+2)} \right|\\&=\dfrac{13n+6}{(n+1)(n+2)}.\end{align}$$ We want to show $\dfrac{13n+6}{(n+1)(n+2)}$ is less than or equal to some simple fraction which can still be made less than $\epsilon$. Since replacing the numerator with a larger value and replacing the denominator with a smaller value both increases the size of the expression respectively, we have $$\dfrac{13n+6}{(n+1)(n+2)}\leq \dfrac{19n}{(n+1)(n+2)}\leq \dfrac{19n}{n^2}=\dfrac{19}{n}.$$ Now to ensure $\dfrac{19}{n}<\epsilon$, we can take $n>\dfrac{19}{\epsilon}$. But we are not done yet. Recall that the definition requires $N$ to be a positive integer. To fix this, we take $N$ to be the integer part or floor of $\dfrac{19}{\epsilon}$, which is defined to be the largest integer less than or equal to $\dfrac{19}{\epsilon}$ and denoted by $\lfloor \dfrac{19}{\epsilon} \rfloor$. We can now proceed to the proof.
Proof: Let $\epsilon>0$ and $N=\lfloor \dfrac{19}{\epsilon}\rfloor \in \Bbb Z^+$. Then for all integers $n>N$, we have $$\begin{align}\left| \dfrac{3n^2-4n}{(n+1)(n+2)}-3 \right|&=\left| \dfrac{3n^2-4n-3(n+1)(n+2)}{(n+1)(n+2)} \right|\\&=\left| \dfrac{3n^2-4n-(3n^2+9n+6)}{(n+1)(n+2)} \right|\\&=\left| \dfrac{-13n-6}{(n+1)(n+2)} \right|\\&=\dfrac{13n+6}{(n+1)(n+2)}\\ &\leq \dfrac{13+6n}{(n+1)(n+2)}\\&\leq \dfrac{19n}{n^2}\\&=\dfrac{19}{n}\\&< \epsilon\quad \text{since}\;n>\dfrac{19}{\epsilon}.\end{align}$$ Therefore $\lim\limits_{n\to \infty}\dfrac{3n^2-4n}{(n+1)(n+2)}=3$.
Limits of functions
Definition: Let $f:A\subset \Bbb R \to B\subset \Bbb R$. Let $a\in A$. Then $f(x)$ tends to a limit $l$ as $x$ tends to $a$ if, for any $\epsilon>0$, there exists $\delta>0$ such that $$0<|x-a|<\delta \Rightarrow |f(x)-l|<\epsilon.$$ We can write either $f(x)\to l$ as $x\to a$ or $\lim\limits_{x\to a}f(x)=l$.
Example
Show that $\lim\limits_{x\to 2}2x^2-5x=-2$.
We need to ensure that $|f(x)-l|=|2x^2-5x+2|$ is less than any specified positive $\epsilon$ by taking $x$ such that $0<|x-2|<\delta$. So we consider $|2x^2-5x+2|$: $$|2x^2-5x+2|=|(x-2)(2x-1)|=|x-2||2x-1|.$$ We can control the size of $|x-2|$ since we can choose $\delta$ such that $|x-2|<\delta$. As for $|2x-1|$, we can rewrite it in terms of $|x-2|$ using the triangle inequality: $$\begin{align}|2x-1|&=|2(x-2)+3|\\ &\leq |2(x-2)|+|3|\\ &=2|x-2|+3.\end{align}$$ Therefore $$|2x^2-5x+2|=|x-2||2x-1|\leq |x-2|(2|x-2|+3).$$ Say $|x-2|<1$. Then $2|x-2|+3<5$. If, in addition, $|x-2|<\dfrac{\epsilon}{5}$, then $|x-2|(2|x-2|+3)$ will be less than $\epsilon$. We are now ready to start the proof.
Proof: Let $\epsilon>0$ and $\delta=\min\{1,\dfrac{\epsilon}{5}\}$. Then $$0<|x-2|<\delta \Rightarrow |x-2|<1 \wedge |x-2|<\frac{\epsilon}{5}\\ \begin{align}\Rightarrow |2x^2-5x+2|&=|x-2||2x-1|\\ &\leq |x-2|(2|x-2|+3)\\ &<5|x-2|\\ &<\epsilon \end{align}$$ Therefore $\lim\limits_{x\to 2}2x^2-5x=-2$.
Remark: The choice of $\delta=\min\{1,\dfrac{\epsilon}{5}\}$ is arbitrary. Choosing $\delta=\min\{\dfrac{1}{2},\dfrac{\epsilon}{4}\}$ will also work.
Convergence of sequences
Definition: The sequence $(a_n)$ converges to $l$ if, for any $\epsilon>0$, there exists $N\in \Bbb Z^+$ such that $$n>N \Rightarrow |a_n-l|<\epsilon.$$ We can write either $a_n\to l$ as $n\to \infty$ or $\lim\limits_{n\to \infty}a_n=l$.
Example
A sequence $(a_n)$ is defined by $a_n=\dfrac{3n^2-4n}{(n+1)(n+2)}$ for $n\in \Bbb Z^+$. Show that $\lim\limits_{n\to \infty}a_n=3$.
The definition of convergence involves two quantifiers and is of the form $$\forall \epsilon\;\exists N \bullet P(\epsilon,N)$$ where $P(\epsilon,N)$ is a propositional function and the universes for $\epsilon$ and $N$ are $\Bbb R^+$ and $\Bbb Z^+$ respectively. The structure of the proof starts with an arbitrary $\epsilon$ (in its universe) and then selects a particular $N$ (in its universe) which may depend on $\epsilon$. To complete the proof, we need to show that the propositional function $P(\epsilon,N)$ is satisfied for the arbitrary $\epsilon$ and particular $N$.
We first start by the definition. We need to find a positive integer $N$ such that for all $n>N$, $$|a_n-l|=\left|\frac{3n^2-4n}{(n+1)(n+2)}-3 \right|<\epsilon.$$ But how? We can manipulate $|a_n-l|$ until it is less than any positive $\epsilon$: $$\begin{align}\left| \dfrac{3n^2-4n}{(n+1)(n+2)}-3 \right|&=\left| \dfrac{3n^2-4n-3(n+1)(n+2)}{(n+1)(n+2)} \right|\\&=\left| \dfrac{3n^2-4n-(3n^2+9n+6)}{(n+1)(n+2)} \right|\\&=\left| \dfrac{-13n-6}{(n+1)(n+2)} \right|\\&=\dfrac{13n+6}{(n+1)(n+2)}.\end{align}$$ We want to show $\dfrac{13n+6}{(n+1)(n+2)}$ is less than or equal to some simple fraction which can still be made less than $\epsilon$. Since replacing the numerator with a larger value and replacing the denominator with a smaller value both increases the size of the expression respectively, we have $$\dfrac{13n+6}{(n+1)(n+2)}\leq \dfrac{19n}{(n+1)(n+2)}\leq \dfrac{19n}{n^2}=\dfrac{19}{n}.$$ Now to ensure $\dfrac{19}{n}<\epsilon$, we can take $n>\dfrac{19}{\epsilon}$. But we are not done yet. Recall that the definition requires $N$ to be a positive integer. To fix this, we take $N$ to be the integer part or floor of $\dfrac{19}{\epsilon}$, which is defined to be the largest integer less than or equal to $\dfrac{19}{\epsilon}$ and denoted by $\lfloor \dfrac{19}{\epsilon} \rfloor$. We can now proceed to the proof.
Proof: Let $\epsilon>0$ and $N=\lfloor \dfrac{19}{\epsilon}\rfloor \in \Bbb Z^+$. Then for all integers $n>N$, we have $$\begin{align}\left| \dfrac{3n^2-4n}{(n+1)(n+2)}-3 \right|&=\left| \dfrac{3n^2-4n-3(n+1)(n+2)}{(n+1)(n+2)} \right|\\&=\left| \dfrac{3n^2-4n-(3n^2+9n+6)}{(n+1)(n+2)} \right|\\&=\left| \dfrac{-13n-6}{(n+1)(n+2)} \right|\\&=\dfrac{13n+6}{(n+1)(n+2)}\\ &\leq \dfrac{13+6n}{(n+1)(n+2)}\\&\leq \dfrac{19n}{n^2}\\&=\dfrac{19}{n}\\&< \epsilon\quad \text{since}\;n>\dfrac{19}{\epsilon}.\end{align}$$ Therefore $\lim\limits_{n\to \infty}\dfrac{3n^2-4n}{(n+1)(n+2)}=3$.
Limits of functions
Definition: Let $f:A\subset \Bbb R \to B\subset \Bbb R$. Let $a\in A$. Then $f(x)$ tends to a limit $l$ as $x$ tends to $a$ if, for any $\epsilon>0$, there exists $\delta>0$ such that $$0<|x-a|<\delta \Rightarrow |f(x)-l|<\epsilon.$$ We can write either $f(x)\to l$ as $x\to a$ or $\lim\limits_{x\to a}f(x)=l$.
Example
Show that $\lim\limits_{x\to 2}2x^2-5x=-2$.
We need to ensure that $|f(x)-l|=|2x^2-5x+2|$ is less than any specified positive $\epsilon$ by taking $x$ such that $0<|x-2|<\delta$. So we consider $|2x^2-5x+2|$: $$|2x^2-5x+2|=|(x-2)(2x-1)|=|x-2||2x-1|.$$ We can control the size of $|x-2|$ since we can choose $\delta$ such that $|x-2|<\delta$. As for $|2x-1|$, we can rewrite it in terms of $|x-2|$ using the triangle inequality: $$\begin{align}|2x-1|&=|2(x-2)+3|\\ &\leq |2(x-2)|+|3|\\ &=2|x-2|+3.\end{align}$$ Therefore $$|2x^2-5x+2|=|x-2||2x-1|\leq |x-2|(2|x-2|+3).$$ Say $|x-2|<1$. Then $2|x-2|+3<5$. If, in addition, $|x-2|<\dfrac{\epsilon}{5}$, then $|x-2|(2|x-2|+3)$ will be less than $\epsilon$. We are now ready to start the proof.
Proof: Let $\epsilon>0$ and $\delta=\min\{1,\dfrac{\epsilon}{5}\}$. Then $$0<|x-2|<\delta \Rightarrow |x-2|<1 \wedge |x-2|<\frac{\epsilon}{5}\\ \begin{align}\Rightarrow |2x^2-5x+2|&=|x-2||2x-1|\\ &\leq |x-2|(2|x-2|+3)\\ &<5|x-2|\\ &<\epsilon \end{align}$$ Therefore $\lim\limits_{x\to 2}2x^2-5x=-2$.
Remark: The choice of $\delta=\min\{1,\dfrac{\epsilon}{5}\}$ is arbitrary. Choosing $\delta=\min\{\dfrac{1}{2},\dfrac{\epsilon}{4}\}$ will also work.
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