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Show with $\varepsilon$-$\delta$ that the function $$f(x) = \frac{2x+3}{5}$$ is continuous.

I found this task on the internet but it was without solution, so I'm asking here if I did it correctly.

Let $\varepsilon>0$ and let $\delta = \frac{\varepsilon}{2}$. If $|x-x_{0}|< \delta$, then:

$$\left |\frac{2x+3}{5} - \left(\frac{2x_{0}+3}{5}\right)\right | = \left|\frac{2x+3-(2x_{0}+3)}{5}\right|=\left|\frac{2x+3-2x_{0}-3}{5}\right|$$

$$=\left|\frac{2x-2x_{0}}{5}\right|<|2x-2x_{0}|=\left|2(x-x_{0})\right|<|2\delta|=\varepsilon$$

Therefor, the function is continuous.


Is it alright?

I'm not sure because I have just removed $5$ from the denominator and claimed it's greater without (it really is..). But am I actually allowed to do that?

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    $\begingroup$ Your proof is correct $\endgroup$ – Sathasivam K Sep 20 '16 at 11:16
  • $\begingroup$ your proof is correct only because of you remove 5from denominator $\endgroup$ – Sathasivam K Sep 20 '16 at 11:20
  • $\begingroup$ Since yo have modulus you can able to remove 5 $\endgroup$ – Sathasivam K Sep 20 '16 at 11:21
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    $\begingroup$ Alternatively you can keep the 5 and let $\delta=5\varepsilon/2$. $\endgroup$ – Olivier Moschetta Sep 20 '16 at 11:26
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    $\begingroup$ Guys I don't get it... Is my proof correct or wrong? $\endgroup$ – cnmesr Sep 20 '16 at 14:55
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Your proof is complete. Are you doubting that whenever $A < B$ and $B < C$, then also $A < C$ ? Don't :)

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  • $\begingroup$ could not comment this. This was more like a comment $\endgroup$ – Sargsyan Grigor Sep 22 '16 at 16:47

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