anonymous
  • anonymous
Prove the limit as x approaches -2 of (x^2-1) = 3 This is the homework hints website but I'm not understanding all it's saying either. Thank you.
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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blurbendy
  • blurbendy
I'm not sure how to go about proving it other than by actually evaluating the limit at x = -2 lim ( (-2)^2 - 1) = 4 - 1 = 3 x -> -2
anonymous
  • anonymous
Thank you. This is some kind of proof where you find the delta along the x axis when given an epsilon along the y axis. Thank you for your time.
anonymous
  • anonymous
So, do you have the epsilon?

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anonymous
  • anonymous
No, I believe they want me to solve for all epsilon > 0.
anonymous
  • anonymous
\[ |x-(-2)|<\delta \implies | (x^2-1) - 3|<\epsilon \]
anonymous
  • anonymous
\[ |x+2|<\delta \implies | x^2-4 |<\epsilon \]
anonymous
  • anonymous
\[ |x+2|<\delta \implies | (x-2)(x+2) |<\epsilon \\ |x+2|<\delta \implies | x-2||x+2 |<\epsilon \]
anonymous
  • anonymous
Basically ideally we'd like to chose: \[ \delta \sim \frac{\epsilon}{|x-2|} \]But we don't get to chose what \(x\) is. They do. We have to be safe for all cases.
anonymous
  • anonymous
Thank you for the detail.
anonymous
  • anonymous
To prevent \(\delta\) from getting to large, we'll say: \[ \delta \sim \min\{1,f(\epsilon)\} \]
anonymous
  • anonymous
Now that we know \(\delta<1\) we can figure out what the range of \(|x-2|\) is.
anonymous
  • anonymous
\[ |x+2|<1\implies -1
anonymous
  • anonymous
Likewise\[ |x-2|<5\implies \frac{\epsilon}{5}< \frac \epsilon {|x-2|} \]
anonymous
  • anonymous
So our final answer to the challenge is: \[ \delta \sim f(\epsilon) =\min\left\{1,\frac \epsilon 5\right\} \]
anonymous
  • anonymous
Finally, this completes the proof.

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