This is hard to follow. Unless you are using notation eccentrically, both of those limits are $1$.

The existence of $a$ would imply that $\lim_{x \to 1^-}x < \lim_{x \to 1}x.$ What does that mean?

sorry i have made changes to rhe question please check again

If limit exists, it's a number. In this case, it's one on both ends. So essentially you have $1 < a < 1$.

I think it should be more clear now

Now you're asking if $\lim_{x \to 1^-}x < 1.$ What do you think that means?

I think that they're is no $a $ inbetween as the limit is supposed to be as close as it can be to $1 $without being $ 1 $ but just want to confirm if my logic holds

On the real number system, there is no such $a$. But in other number systems, who knows? First you would have to tell us what number system you are using.

i am using the real number system thanks for the answer i will check mark it if u post it as solution

Hint: Assume there is such an $a$. Apply the $\varepsilon, \delta$ definition for $\lim\limits_{x \to 1^-}x$, using $\varepsilon = 1-a$.

For some related thoughts, though, you might be interested in the hyperreal numbers.

thanks for the info

No, $\lim_\limits{x\to 1^-} x = 1.$ It doesn't equal something almost equal to but slightly less than one, it equals $1$ exactly. There is no room to fit a $a$ between $1$ and $1.$ It is a little bit like how $0.\bar9 = 1.$ Exactly $1,$ and not a real number that is "rigth next to" $1.$

No, that limit equals $1$, so it doesn't make sense to say $1<a<1$.

@Accelerator so lets say that the max $(f(x)) = 1 $ and $g(x) = \lim\limits_{g(x) \to 1^-}g(x)$ then is it fair to say that $g(x) \geq f(x)$ when $f(x) \neq 1$?

@Accelerator g(x) is basicaly just g(x) = 1 but smaller than 1 (although i dont think concept is correct anymore)

I don't know if you're just making it more complicated than it has to be, but that still doesn't make sense. If you say out loud "$g(x)$ equals $1$ but smaller than $1$", you can see what I mean.

Because $f(x)=x$ is continuous at $1,$ we know that $\lim_{x\to1}x=1.$ The fact that the limit exists also implies that $\lim_{x\to1^-}x=\lim_{x\to1}x=\lim_{x\to1^+}x.$ That is, all three limits are exactly equal to $1.$ It is not true that the lower limit is "as close as it can be to 1 without being 1" -- the limit is $1$. In fact there is nothing "as close as it can be to 1 without being 1"; if you claimed to have found such a number, there would be another number $a$ between your number and $1.$