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Prove that the function $\displaystyle f(x)=\frac{x}{x^2+1}$ is continuous at $x=-1$ using the definition of continuity.

My attempt:

Need show that $\forall {\epsilon > 0},\exists {\delta > 0}$ s.t $\forall {x \in R },\ |x - (-1)| < \delta \implies |f(x) - f(-1)| < \epsilon.\\ |f(x) - f(-1)| = |\frac{x}{x^2 + 1} - \frac{-1}{2}| = |\frac{2x - (-1)(x^2 + 1)}{2(x^2 + 1)}| = |\frac{2x + x^2 + 1)}{2(x^2 + 1)}| = |\frac{(x+1)^2}{2(x^2 + 1)}|\\ |\frac{(x+1)^2}{2(x^2 + 1)}| < |\frac{(x+1)^2}{(x^2 + 1)}| < |(x+1)^2| \text{ since } (x^2 + 1) \text{ is at least 1}\\ |(x+1)^2| = |x + 1|^2 < \epsilon \\ |x + 1| < \sqrt{\epsilon}$

Therefore set $\delta = \sqrt{\epsilon} \text{ and } \forall {\epsilon > 0}, \ \forall {x \in R }, |x - (-1)| < \sqrt{\epsilon} \implies |f(x) - f(-1)| < \epsilon$

is this a valid way of proving this? Or is there a better way?

Prabath Hewasundara
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Carlisle Manson
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2 Answers2

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Your proof and choice of $\delta$ are OK. There's one minor issue. The inequalities below are non-strict since you have equality when $x=-1$: $$\left|\frac{\left(x+1\right)^2}{2\left(x^2+1\right)}\right|\leq \left|\frac{\left(x+1\right)^2}{x^2+1}\right|\leq\left|\left(x+1\right)^2\right| $$ This doesn't change the rest of the proof.

bjorn93
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  • Strictly, the epsilon-delta definition should have $\textbf{0}<|x-c|<\delta$, so $x$ shouldn't be able to take the value of $x=-1$ (Question 958412). But indeed, the OP should have included that in the definition. – Jam Feb 02 '20 at 21:56
  • @Jam No, OP is proving continuity using the $\epsilon$-$\delta$ definition of it, not the limit definition. So $x$ can be $-1$. – bjorn93 Feb 02 '20 at 21:58
  • @Jam see the link for the $\epsilon$-$\delta$ definition: https://en.wikipedia.org/wiki/Continuous_function#Weierstrass_and_Jordan_definitions_(epsilon%E2%80%93delta)_of_continuous_functions – bjorn93 Feb 02 '20 at 22:06
  • The value of a function at a point doesn't affect the limit of the function at the point. So, we would only take $x=-1$ when examining the value of the function, not the limit. If you allow $0\color{red}{=}|x-(-1)|<\delta$, the epsilon-delta definition of the limit breaks down, e.g., $f(x)=\begin{cases}1,&x=-1\\text{undefined},&\text{else}\end{cases}$ would have a "limit" at $x=-1$ since we can always take $|x-(-1)|<\delta$ despite only being defined at one point. – Jam Feb 02 '20 at 22:08
  • @Jam Yes but that applies when we're taking limits. We're not doing this here. OP is using a direct $\epsilon$-$\delta$ definition of continuity which is cited in the link in my previous comment. Since $f(-1)$ has to be defined, there's no need to impose the $0<|x-(-1)|$ requirement. Of course, if you want, you can prove that the limit at $-1$ is $-1/2$ in which case we'd have that requirement. It's equivalent, really. – bjorn93 Feb 02 '20 at 22:18
  • I didn't realise there were different definitions. Thanks for enlightening me. – Jam Feb 02 '20 at 22:33
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The proof looks correct to me. And I think it's about as neat and direct a proof as you could get.

However, in the $\varepsilon$-$\delta$ definition of the limit, we should have $\textbf{0}<|x-c|<\delta$, since we are interested in the behaviour around the point, not at the point. See (Question 958412). Alternatively, as bjorn93 has pointed out, you could change your strict inequalities to non-strict ones.

Jam
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