The original function is:
$$f(x)=\frac{\sin(x+1)}{x+1}$$
If I wanted to use u-substitution to prove that the function has a removable discontinuity at $x=-1$ would I start out by writing this:
$$f(u-1)=\frac{\sin(u)}{u}$$
Then I using limits I write $\lim_\limits{x \to u-1}f(x)=\frac{\sin(u)}{u}$, then I would take the limit of that as u approaches 0 getting $\lim_\limits{u \to 0}\frac{\sin(u)}{u}=1$. Therefore I can say that the first limit is equal to the second limit hence the function has a removable discontinuity at $x=-1$. Is the way I proved the removable discontinuity precise, or is there fallacy in my thinking?
The reasoning is correct, but nowhere do I see you actually proving that the limit is $1$. Have you done that?