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Show that the function $ f(x)=\frac{1}{x+1}$ is continuous at $x=1$ using the $\varepsilon$-$\delta$ definition of continuity.

My initial thoughts were that this is a straight plug in of value at $x=1$, which would give me $f(1)=\frac{1}{1+1}$ .

Missing some crucial concepts?

Zev Chonoles
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Sylvester
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    It says you need to use the $\epsilon-\delta$ definition; do you know the definition? –  Feb 03 '14 at 03:29
  • The definition of continuity of functions on $\mathbb{R}$ tells us that the limit of any continuous function as $x$ approaches $a$ is precisely the value of the function at $a$.

    Plugging in $x=1$ won't prove it - what if there was a sudden jump at $f(1)$? So, as T Bongers said, you need to use the $\epsilon - \delta$ definition.

     – Lost Feb 03 '14 at 03:36
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    Ahem, this is a pet peeve of mine: it is the $\epsilon$ hyphen $\delta$ definition, not the $\epsilon$ minus $\delta$ definition. So please don't use math mode where it doesn't belong. – Zev Chonoles Feb 03 '14 at 03:37
  • you gotta see how to make $|f(x)-f(1)|$ small as you wish, that is, less than any given $\epsilon>0$, by taking $|x-1|$ also small enough, $\delta$. This $\delta$ depends on $\epsilon$. – janmarqz Feb 03 '14 at 03:45
  • A related technique. – Mhenni Benghorbal Feb 03 '14 at 04:01

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$$ |f(x) - f(1)| = |\frac {1} {x+1}- \frac 1 2| = |\frac {x - 1}{2(x + 1)}| \le \frac 1 2|x - 1|$$

This implies that for any given positive value $\epsilon$ there exists a positive value $\delta = 2\epsilon$ such that $|f(x) - f(1)| \le \epsilon$ whenever $|x - 1| \le \delta$.

This says that the limit of $f(x)$ as $x$ approaches $1$ is $f(1)$ which implies that $f$ is continuous at $x = 1$.

This says no matter how small $\epsilon $ is there corresponds another positive value $\delta$ such that for all values of $x$ that are at most $\delta$ distant from $1$, the values of the function $f(x)$ will be at most $\epsilon $ distant from $f(1)$.

No matter how close you want $f(x)$ to be to $f(1)$ there are arbitrarily many values of $x$ close to 1.

Intuitively there is no break of the sequence of the values of the function either side of $x = 1$.

None of this matters of course if you have not grasped the $\epsilon $ - $\delta$ definition for continuity.

If you want to read about it try "Differential Calculus" by Shanti Narayan or "Calculus" by Michael Spivak. Might want to take a look at this video too..

Ishfaaq
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