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 $-\infty$?
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.