Question

Use the epsilon-delta definition of limit to pro that:
lim (x^2) as (x approaches 2) = 4

Answer 1

\[\lim_{x \rightarrow 2}x^2=4\\ \forall \in >0 \ \exists \delta >0: |x-2|<\delta \rightarrow |x^2-4|<\in\\|x^2-4|=|x-2||x+2|<\in \\|x-2|<\delta \rightarrow |x|<2+\delta \\max \ \delta <\frac{ 1 }{ 2 }\rightarrow |x|<2+\frac{ 1 }{ 2}\rightarrow |x+2|<\frac{ 5 }{ 2}\\|x-2||x+2|<\in \rightarrow |x-2|\frac{5}{2}<\in\\|x-2|<\frac{2}{5} \in\\now \\ \delta<\max \ (\frac{1}{2},\frac{2}{5} \in)\]

Answer 2

in the last line must be
\[\delta < \min ( \frac{ 1 }{ 2},\frac{ 5 \in }{ 2} )\]

Answer 3

@amoodarya should that be

$$
\delta {\color{red}=} \min ( \frac{ 1 }{ 2},\frac{ 5 \in }{ 2} )
$$

Answer 4

@ybarrap
= is ok
but < is more reliable
like this
|x-1|<.01 ---> |x-1|<.5
now suppose |x-1|<delta
and delta <0.1 so
|x-1|<delta --->|x-1|<0.5

Answer 5

that makes sense, thanks