# Prove that $\lim_{x \to a }cx^{2}+dx+e=ca^{2}+da+e$

I will prove by use properties of limit

$$\lim_{x \to a }cx^{2}+dx+e=\lim_{x \to a}cx^{2}+\lim_{x \to a}dx+\lim_{x\to a }e$$

we begin $\lim_{x \to a} cx^{2}$

Given $\epsilon ,\frac{\epsilon }{3},\delta _{1}>0$

$$0<|x-a|<\delta _{1}\rightarrow |cx^{2}-ca^{2}<\frac{\epsilon }{3}$$ $$\rightarrow c|x^{2}-a^{2}|<\frac{\epsilon }{3}$$ $$\rightarrow |x^{2}-a^{2}|<\frac{\epsilon }{3c}$$

Choose $\delta _{1}=\frac{\epsilon }{3c}$

$$|x^{2}-a^{2}|<c\frac{\epsilon }{3c}=\frac{\epsilon }{3}$$

$\lim_{x \to a }dx=da|$ case

given $\epsilon ,\frac{\epsilon }{3},\delta _{2}>0$

$$0<|x-a|<\delta _{2}\rightarrow |dx-da|< \frac{\epsilon }{3}$$ $$\rightarrow d|x-a|<\frac{\epsilon }{3}$$ $$|x-a|<\frac{\epsilon }{3d}$$

Choose $\delta _{2}=\frac{\epsilon }{3d}$

then

$$|dx-da|<d\frac{\epsilon }{3d}=\frac{\epsilon }{3}$$

$\lim_{x \to a}e$

given $\epsilon ,\frac{\epsilon }{3},\delta _{3}$

$0<|x-a|<_{3}\rightarrow |e-e|=0<\frac{\epsilon }{3}$

Combine all limit

$\lim_{x \to a}cx^{2}+\lim_{x \to a}dx+\lim_{x\to a }e$

$|cx^{2}-ca^{2}|+|dx-da|+|e-e|<\frac{\epsilon }{3}+\frac{\epsilon }{3}+\frac{\epsilon }{3}=\epsilon$

• You only need to use the result which says that if $P(x)$ is polynomial and $c\in\Bbb R$ then $$\lim_{x\to c}P(x)=P(c).$$ Commented Apr 4, 2017 at 11:26
• Are you required to prove the problem using $\epsilon$-$\delta$ definition? Commented Apr 4, 2017 at 11:37
• yes because I have no idea to prove by other method Commented Apr 4, 2017 at 11:39
• The first part of your proof is not correct. You need to factor $x^2-a^2$ as $(x-a)(x+a)$. Commented Apr 4, 2017 at 11:42

For the first case you have to be a bit careful. First of all, let's assume that $c \neq 0$ (or else it's trivial to prove it. We have that:

$0 < \vert x-a \vert < \delta_1$ (for some $\delta_1$ soon to be determined)

Now, let's agree to pick a $\delta_1$ such that $\delta_1 \leq 1$. This will help us later as $0 < \vert x-a \vert < \delta_1 \leq 1$ now implies that:

$$\vert x \vert - \vert a \vert \leq \vert x - a \vert < 1 \Rightarrow \vert x \vert < 1 + \vert a \vert$$

From here, if we pick:

$$\delta_1 = \min \left(1, \frac{\epsilon}{c(1 + 2\vert a \vert)}\right)$$

(Note: don't let the "min" part throw you off, we are essentially ensuring that $\delta_1$ is smaller than both quantities)

we get that:

$$\vert cx^2 - ca^2 \vert = c\vert x-a \vert \vert x+a\vert \leq c\vert x-a \vert (\vert x\vert + \vert a \vert) < c\vert x-a \vert (1 + 2\vert a \vert) < c\delta_1 (1 + 2\vert a \vert) < \epsilon$$

as desired.