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anyone can help me with this question? I know that to prove continuity the left side limit must equal the right side limit, in this case is 3. But there is only one equation for me to do so, so any idea how to solve this question?

Is it possible to choose a value less than 3 and a value more than 3 and sub it into the equation? If that’s the case the limit will not be the same.

Really confused with this question, any help is appreciated. Thank you.

Raffaele
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user754578
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  • $\cdots$ in this case "at $x=3$." – mjw Feb 16 '21 at 10:53
  • The limit is not the value at a point greater than $3$ or at a point less than $3$ - the limit is what happens when you get close to $3$. The question is "can I make $f(x)$ as close as I like to $f(3)$ by making sure that $x$ is close enough to $3$" - so you simply need to show you can make $|f(x)-f(3)|$ as small as you like by making $|x-3|$ small enough. – Mark Bennet Feb 16 '21 at 11:00

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We are interested in the behavior of $f$ "near" the point $3$, hence we can assume that $2 <x <4.$

Then we have

$$|f(x)-f(3)|=|x^2-9|=|x-3| \cdot |x+3|=|x-3|(x+3) \le 7|x-3|.$$

Can you proceed with $ \varepsilon , \delta$ ?

Fred
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  • Thanks for your help. I have come out with this for any epsilon more than 0, As Long as 0<|x-3|<delta, |f(x)-3|<|x-3+6||x-3|<something=epsilon. I cant figure out what value to put for delta. – user754578 Feb 16 '21 at 18:12
  • I have try using delta = (-3sqrt(6+epsilon) ) and for the something i put delta^2 - 5epsilon but cannot seems to get just the epsilon value. – user754578 Feb 16 '21 at 18:15
  • $7|x-3| < \epsilon \iff |x-3|<\frac{7}{\epsilon}.$ Hence $\delta = \frac{7}{\epsilon}$ is a good idea. – Fred Feb 17 '21 at 08:02
  • Can i do it like this, since delta = 7/ epsilon, by the definition, as Long as 0<|x-3|<delta, then |f(x)-f(3)|=|x^2-9|=|x-3||x+3|<7|x-3|<epsilon/7=epsilon. Therefore, the function is continuous at x=3. – user754578 Feb 17 '21 at 09:44