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Okay, so I've been asked to prove this limit using said epsilon-delta definition:

$ \lim_{x \rightarrow 7} \sqrt{x-3} =2$

I have gotten thus far:

Let epsilon > $0$. Then if delta=_______, then $|x-7|<delta.$

$|f(x)-L|= | \sqrt{x-3} -2|$

$ = |\frac{( \sqrt{x-3}-2)( \sqrt{x-3} +2)}{ \sqrt{x-3} +2} |$

$= |\frac{x-7}{ \sqrt{x-3}+2 } |$

$= \frac{1}{ \sqrt{x-3} +2} |x-7|$

and am now stuck. How do I solve for delta now?

Thanks in advance.

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  • $\begingroup$ Since you have the basics of mathjax mastered, please format you unfinished proof, directly into your question. Please do not use images, but rather take the time to format the entire post. $\endgroup$
    – amWhy
    Jan 31, 2022 at 0:28
  • $\begingroup$ Done. Sorry about that. $\endgroup$ Jan 31, 2022 at 0:46
  • $\begingroup$ $\lim_{x\to 7} \sqrt{x-3}=2$ $\endgroup$ Jan 31, 2022 at 0:50
  • $\begingroup$ @eleganthaunt Nice work! Thanks. $\endgroup$
    – amWhy
    Jan 31, 2022 at 0:53

1 Answer 1

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You want to conclude $\frac{|x-7|}{\sqrt{x-3}+2}<\epsilon$ and you can control $|x-7|<\delta$.

We can force $|x-7|\leq 1$ from which follows $x-7\geq -1$ so that $x-3\geq 3$ so that the square root is well defined. Therefore the denominator is bigger than $1$ and so

$$\frac{|x-7|}{\sqrt{x-3}+2}<|x-7|$$

So it is sufficient to pick $\delta\leq \epsilon$. We need to force these two conditions together, so given $\epsilon>0$ we let $\delta=\min(\epsilon, 1)$.

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  • $\begingroup$ Thank you! I edited that the limit was =2 instead of to 7, and you saw that, right? Bad copy error on my part $\endgroup$ Jan 31, 2022 at 1:10
  • $\begingroup$ @eleganthaunt Yes, I picked up from your $\frac{|x-7|}{\sqrt{x-3}+2}$ which seems correct. $\endgroup$
    – Snaw
    Jan 31, 2022 at 1:16

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