Answer
$$\lim_{x\to1^-}\Big(\frac{1}{x+1}\Big)\Big(\frac{x+6}{x}\Big)\Big(\frac{3-x}{7}\Big)=1$$
Work Step by Step
$$\lim_{x\to1^-}\Big(\frac{1}{x+1}\Big)\Big(\frac{x+6}{x}\Big)\Big(\frac{3-x}{7}\Big)$$
To find one-side limit algebraically, we still apply the limit laws like for the two-side limits as normal.
$$A=\lim_{x\to1^-}\Big(\frac{1}{x+1}\Big)\Big(\frac{x+6}{x}\Big)\Big(\frac{3-x}{7}\Big)$$ $$A=\Big(\frac{1}{1+1}\Big)\Big(\frac{1+6}{1}\Big)\Big(\frac{3-1}{7}\Big)$$ $$A=\frac{1}{2}\times\frac{7}{1}\times\frac{2}{7}$$ $$A=\frac{1}{1}=1$$
Therefore, $$\lim_{x\to1^-}\Big(\frac{1}{x+1}\Big)\Big(\frac{x+6}{x}\Big)\Big(\frac{3-x}{7}\Big)=1$$
