If a function $f(x)$ is continuous at $x=a$, then $\lim_{x\to a}f(x) = f(a).$
If $f(x)$ is continuous at $a$, and $g(x)$ is continuous at $f(a)$, then $\lim_{x\to a}g(f(x)) = g(f(a)).$
Here you have the function $f(x) = x^2-x$, which is continuous everywhere, and the function $g(x) = e^x$, which is continuous everywhere. So you can compute the limit by evaluation.
Your argument is upside down: you are asked to use continuity to evaluate the limit, but you are trying to conclude that the function is continuous by arguing that the value of the limit equals the value of the function. Not only is that not what you were asked to do, but more importantly: how did you verify that the value of the limit was indeed $1$, without using continuity?