How to prove that $f(x)=\frac {\sin(1+x)}{1+x}\;(x \neq 1),\; 1 \;(x=1)$ is continuous? I need to prove that the following function is continuous:
$$
f(x)=
\begin{cases}
  \dfrac {\sin(1+x)}{1+x}&,\;x \neq -1 \\
  1&,\;x=-1
\end{cases}$$
 A: Hint: let $u = 1+x$. Then,
$$\lim_{x\to -1}\frac{\sin(1+x)}{1+x} = \lim_{u\to 0}\frac{\sin(u)}{u}$$
A: It is in fact a real analytic function:
$$f(x) = \sum_{n\ge 0} (-1)^n \frac{(x+1)^{2n}}{(2n+1)!} = 1 - \frac{(x+1)^2}{6} + \frac{(x+1)^4}{120} - \cdots  $$
for all $x$. 
A: First let F be a function such that I=[a,b], I ⊆ F: ∃ x0 ∈ I, a < x0 < b, ∃ lim f(x), x->x0.
So we have that:
$$ \exists \lim_{x \to x_{0}}f(x): \lim_{x \to x_{0}}(f(x) - f(x_{0})) = 0 $$
And:
$$ \exists \lim_{x \to x_{0}^{+}}f(x): \lim_{x \to x_{0}^{+}}f(x) = f(x_{0}) \wedge \exists \lim_{x \to x_{0}^{-}}f(x): \lim_{x \to x_{0}^{-}}f(x) = f(x_{0}) $$
So, we can colclude that a function F is continuos if there is a x0 such that:
$$ \lim_{x \to x_{0}^{+}}f(x) = \lim_{x \to x_{0}^{-}}f(x) = \lim_{x \to x_{0}}f(x) = f(x_{0}) $$
If your function fulfills these conditions, then it is continuos. So you just have to make sure that the following is true:
$$ \lim_{x \to -1^{+}} \frac{sin(1+x)}{1+x} = \lim_{x \to -1^{-}} \frac{sin(1+x)}{1+x} = f(-1) = 1 $$
I hope I have helped.
With the best regards,
Saclyr.
