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$$f(x)= \begin{cases} x ; x\le 1 \\ 5;x \gt 1 \end{cases}$$

For example, if I have to check the continuity at $x=0$, I'd simply do,

$$f(0)=0$$

also,$$\lim_{x \to 0}x=0=f(0)$$

Hence, $f$ is continuous at $x=0$.

But the solution manual does something like,

putting $x=x-h$ as $x \to 0^-, h \to 0$

$$\lim_{h \to 0}(0-h)=0$$

and putting $x=x+h$ as $x\to 0^+$ , $h\to 0$

$$\lim_{h\to 0}(0+h)=0$$

Hence, L.H.L.=R.H.L.=$f(0)$ $f$ is continuous at $x=0$.

I think doing all this was rather unnecessary. You don't even need to find the L.H.L. and R.H.L. separately for this case ($x=0$). Or do I? I think I'd only be required to do so if I had to check the continuity at $x=1$, and even in that case I would not assume $x=x-h$ or $x=x+h$, I would have calculated the limit directly.

My question being, is assuming $x=x-h$ or $x=x+h$ for finding one sided limits, as done above, necessary or is it just trivial? Also are there any cases in which the assumption is useful or necessary?

Raknos13
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  • @PiyushDivyanakar Have you read my question? I mentioned that at the end. – Raknos13 Dec 28 '17 at 06:09
  • I always suggest to continuity askers that they include the definition of continuity, often they have an intuitive picture but forget something, nothing wrong with rehashing definition of continuity. Also limit definition. Which continuity depends on, and particularly epsilon and Delta versions. – marshal craft Dec 28 '17 at 07:17
  • @marshalcraft , according to what I've learned, A function $f$ is continuous at point $c$ in its domain, if $\lim_{x \to c}f(x)=f(c)$ or If you can draw the graph of the function without lifting the pen. – Raknos13 Dec 28 '17 at 09:45

1 Answers1

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The difference in

  1. $\lim_{x\to0}f(x)$ and
  2. $\lim_{x\to1}f(x)$

in this case is that $f(x)=x$ on an open interval [eg. $(-1,1)$] containing $0$ but not on an open interval containing $1$.

Thus it is unnecessary to take a two-sided limit at $x=0$, but necessary at $x=1$.

So

$$ \lim_{x\to0}f(x)=\lim_{x\to0}f(x)=\lim_{x\to0}x=0 $$

But

$$ \lim_{x\to1^-}f(x)=\lim_{x\to1^-}x=1 $$

and

$$ \lim_{x\to1^+}f(x)=\lim_{x\to1^+}5=5 $$

so $\lim_{x\to1}f(x)$ does not exist. So $f$ is not continuous at $x=1$.

John Wayland Bales
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