Answer
The limit is
$$\frac{\lim_{(x,y)\to(a,b)}(f(x,y)+g(x,y))}{\lim_{(x,y)\to(a,b)}f(x,y)}=\frac{7}{4}.$$
Work Step by Step
It is given that
$$\lim_{(x,y)\to(a,b)}f(x,y)=4,\quad \lim_{(x,y)\to(a,b)}g(x,y)=3.$$
We will first use the rule that the limit of the quotient is the quotient of the limits:
$$L=\lim_{(x,y)\to(a,b)}\frac{f(x,y)+g(x,y)}{f(x,y)}=\frac{\lim_{(x,y)\to(a,b)}(f(x,y)+g(x,y))}{\lim_{(x,y)\to(a,b)}f(x,y)}.$$
Now we will use the rule that the limit of the sum is the sum of the limits in the numerator:
$$L=\frac{\lim_{(x,y)\to(a,b)}(f(x,y)+g(x,y))}{\lim_{(x,y)\to(a,b)}f(x,y)}=\\\frac{\lim_{(x,y)\to(a,b)}f(x,y)+\lim_{(x,y)\to(a,b)}g(x,y)}{\lim_{(x,y)\to(a,b)}f(x,y)} = \frac{4+3}{4}=\frac{7}{4}.$$
