Proving the Limit of x^2 + 5x - 2 as x Approaches 2 Using Epsilon-Delta Proof

  • Thread starter Thread starter nietzsche
  • Start date Start date
  • Tags Tags
    Proof
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
5 replies · 2K views
nietzsche
Messages
185
Reaction score
0

Homework Statement



Prove that

[tex] \begin{equation*}<br /> \lim_{x \to 2} x^2 + 5x -2 = 12<br /> \end{equation*}[/tex]

Homework Equations





The Attempt at a Solution



We want to prove that given [tex]\varepsilon > 0[/tex], there exists a [tex]\delta[/tex] such that

[tex] 0<|x-2|<\delta \Rightarrow |f(x) - 12| < \varepsilon[/tex]

[tex] \begin{equation*}<br /> f(x)-12\\<br /> = x^2+5x-2-12\\<br /> = (x+7)(x-2)<br /> \end{equation*}[/tex]

So I have an (x-2) term there in the epsilon part. I don't know how to apply that information so that I can choose a delta. Suggestions please!
 
Physics news on Phys.org
sorry, i don't follow

so when i write x+7 as (x-2)+9 i get

f(x) - 12
= (x-2)^2 + 9(x-2)

and it looks like it might be useful, but i don't know how to use it.
 
can you use the fact that the the limit distributes of addition and products?
 
i'm sorry I'm still confused. i have no idea where to go.
 
never mind, i think i figured it out, but not with factoring it like that.

i got [tex] \delta = min(1,\frac{\varepsilon}{10})[/tex]