Prove using the precise definition of the limit

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Homework Statement



lim x→2 (x2+1) = 5

Homework Equations



0 < |x-a| < delta

|f(x) - L| < ε

The Attempt at a Solution



0 < |x-2| < delta

|x2-4| < ε

|(x-2) (x+2)| < ε

|x-2| |x+2| < ε.....and I am stuck here, any help
 
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TheRedDevil18 said:

Homework Statement



lim x→2 (x2+1) = 5

Homework Equations



0 < |x-a| < delta

|f(x) - L| < ε

The Attempt at a Solution



0 < |x-2| < delta

|x2-4| < ε

|(x-2) (x+2)| < ε

|x-2| |x+2| < ε.....and I am stuck here, any help

Continuing your analysis:$$
|x-2|<\frac \epsilon {|x+2|}$$You are trying to get the right side small by getting ##x## close to ##2##. So you don't want the denominator to get close to ##0##. Can you keep it away from ##0## if ##x## is close to ##2##? That's the idea you need to pursue.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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