# What is the limit of a removable discontinuity?

For clarification, consider the function $f \left(x\right) = \frac{\sin \left(x\right)}{x}$. It is clear that there will be some form of a discontinuity at $x = 1$ (as there the denominator is 0). L'hospital's Rule states, in part, that for a function $f \left(x\right) = g \frac{x}{h \left(x\right)}$, where both are differentiable and $h \left(x\right) \ne 0$ on an open interval containing $c$ (except possibly at $c$ itself), if the limits of both functions approach 0 or if both of the functions approach $\pm \infty$ as $x \to c$, the following can be stated.
${\lim}_{x \to a} \frac{g \left(x\right)}{h \left(x\right)} = {\lim}_{x \to a} \frac{g ' \left(x\right)}{h ' \left(x\right)}$
In this case, using L'hospital's Rule with $g \left(x\right) = \sin \left(x\right) , h \left(x\right) = x$ yields $g ' \left(x\right) = \cos \left(x\right) , h ' \left(x\right) = 1.$ Thus, the limit of our original $f \left(x\right)$ as $x \to 0$ will simply be $\cos \frac{0}{1}$, or 1. Notice that if the function were not discontinuous at $x = 0$, this is exactly the value that $f \left(x\right)$ would have taken.