Answer
$$(f \circ g)(x)=(\sqrt{x})^2+3=x+3$$
$$\lim_{x \to 1 }(f \circ g)(x)=4$$
Work Step by Step
To find $\lim_{x\to 1 }(f \circ g)(x)$, examine the graph of $f \circ g$ near $x=1$. As $x$ gets closer to $1$, the values of $(f \circ g)(x)$ get closer to $4$. We conclude from the graph that$$\lim_{x \to 1 }(f \circ g)(x)=4.$$
