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Can you verify the solution to the following (very simple) problem $$\lim_{x\to 1} \frac{x\cdot \sin\{x\}}{x-1},$$ where $\{\}$ is the fractional part function.

LHL = $\lim_{h \to 0^+}\frac{(1-h)\cdot \sin(1-h)}{-h} = \lim_{h \to 0^+}\frac{(1)\cdot \sin(1)}{-h}=-\infty$

RHL = $\lim_{h \to 0^+}\frac{(1+h)\cdot \sin\{1+h\}}{h} = \lim_{h \to 0^+}\frac{(1)\cdot \sin(h)}{h}=1$

Since $LHL \neq RHL$ hence limit does not exist.

Is it correct to say that LHL does not exist? Or should I say LHL exists and is $-\infty$?

Gary
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abhishek
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    If the definition you’re using for the limit of a function (like the one for real functions) does not allow $-\infty$ to be a limit, then you should say that the left-hand limit doesn’t exist. If it does allow it (like the one for extended real functions), then you can say it is $-\infty$. – shoteyes Jun 10 '21 at 07:32
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    It looks good, though the notation is a bit sloppy on the first it is valid. You'd expect there to be no limit here since there's a jump discontinuity on the fractional function at 1 – Alan Jun 10 '21 at 07:35
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    Your derivation gives the right answers, though you give the feeling that at places you are plugging $1$ for $1-h$ inside the limit, which is not rigorous. –  Jun 10 '21 at 08:51

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Since I cannot add comments, here's a short answer: It depends on how you define a limit. A standard definition of the limit of a function is—

A real number $L$ is the limit of a function $f$ at $a \Leftrightarrow \forall \epsilon>0, \, \exists \delta >0$ such that $0<|x-a|< \delta \Rightarrow |f(x)-L|<\epsilon$

If perhaps, you don't understand whatever the bunch of symbols I've written above mean, that's fine. You don't need to. But the important part is that I have defined the limit $L$ to be a real number but $\infty$ is not a real number.

So it suffices to say that "Limit does not exist at $a$" to mean that "There exists no real number $L$ that satisfies those bunch of symbols". However, if you prefer to be more specific, you may add "... because $f(x)\to ∞$ as $x \to a$" after it.

Ibrahim
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