## Calculus (3rd Edition)

$-\infty.$
We have $$\lim _{x \rightarrow 1^-} \frac{3-x}{x-1}= -\infty.$$ We know that the limit from the left will be negative infinity because the numerator will be positive (since 3 is greater than 1), while the denominator will be negative and arbitrarily close to zero.