Proving limit by definition

[Guest]

Hi, can anybody help me with these two limits? I have to prove them by definition of limit. Thank you.

[Guest]

The definition of "limit" is "f(x) has limit L, as x goes to a", [tex]\lim_{x\to a} f(x)= L[/tex], if and only if, given any [tex]\epsilon> 0[/tex] there exist [tex]\delta> 0[/tex] such that if [tex]|x- a|< \delta[/tex] then [tex]|f(x)- f(a)|< \epsilon[/tex].

The second problem asks us to show that [tex]\lim_{x\to -1} 1+ x^2= 2[/tex]. Here, a= -1, [tex]f(x)= 1+ x^2[/tex], and L= 2. Instead of using the definition of "limit" directly, I am going to work "backwards" to see what we need to make [tex]\delta[/tex].

[tex]|f(x)= L|= |1+ x^2- 2|= |x^2- 1|= |(x- 1)(x+ 1)|= |(x+ (-1))(x- (-1))|[/tex]. First, we are going to take x "close to -1" so we can say that -2< x< 0. Then -3< x+ (-1)< -1 and |x+(-1)|= |x-1|< 3. So [tex]|x^2- 1|< 3|x-(-1)|< \epsilon[/tex]. Then [tex]|x-(-1)|< \frac{\epsilon}{3}[/tex] so we can take [tex]\delta= \frac{\epsilon}{3}[/tex].

(Most teachers will accept that as a "proof" but note that the actual proof would go the other way, starting with [tex]|x-(-1)|< \delta= \frac{\epsilon}{3}[/tex] and going to the conclusion that [tex]|(1+ x^2)- 2|< \epsilon[/tex].)