Is $\lim_{x \to -a^{-}}{f(x)}$ the same as $\lim_{x \to (-a)^{+}}{f(x)}$?

Question

Is the limit of $f(x)$ as $x$ appraches $-a$ from the left. $$\lim_{x \to (-a)^{-}}{f(x)}$$ the same as $$\lim_{x \to -a^{-}}{f(x)}$$ where $a>0$, or does removing the parentheses change the meaning?

I reasoned that they are not the same because in the second case, we can multiply $x$ and $-a^{-}$ by $-1$: $$\lim_{-x \to a^{-}}{f(x)}$$ If we substitute in the precise definition of limits this means, $$a-\delta<-x<a\overset{\times -1}{\implies} -a+\delta>x>-a\overset{rearrange}{\implies} (-a)<x<(-a)+\delta$$ which is equivalent to $$\lim_{x \to (-a)^{+}}{f(x)}$$ so $$\lim_{x \to -a^{-}}{f(x)}=\lim_{x \to (-a)^{+}}{f(x)}$$

Answer 1

Thanks to Paramanand Singh's comment, I know what's the fault. Here's is his comment.

"your fault lies in multiplying by $-1$. It should lead to $-x\to a^-$ even if we allow expression in place of variable."

Answer 2

Is the $\lim _{x\to -a^-} f(x)$ the same as $\lim _{x\to (-a)^-} f(x)$?

Yes this is true Both are the limit as $x$ approaches $-a$ from the left.