# Question #af14f

Oct 18, 2015

For any real $a$, ${\lim}_{x \rightarrow a} \left(2 \sin x - 1\right) = 2 \sin a - 1$

#### Explanation:

Additionally, that function has no limits at infinity.

${\lim}_{x \rightarrow - \infty} \left(2 \sin x - 1\right)$ does not exist

and

${\lim}_{x \rightarrow \infty} \left(2 \sin x - 1\right)$ does not exist.