Functions can be combined using algebra, creating new functions.
How can we just add 2 separate functions?
If $f$ and $g$ are functions, we can create a new function $h$ by the rule $h(x) = f(x)+g(x)$. We might call this new function $f+g$.
[Note: Before we evaluate $f(x)$ and $g(x)$, we do have to assume that $x$ is in both the domain of $f$ and that of $g$. But I won't dwell on that because I don't think it's what you're confused about.]
A natural question is: If $f$ and $g$ each have a limit at a point $c$, does $h$? And would that limit depend on the limits of $f$ and $g$?
If you have an intuitive understanding of limit, you can work through this: If values of $f$ are close to $3$ (for instance), and values of $g$ are close to $7$, then values of $h$ ought to be close to $10$. To be a bit more precise, if values of $f$ can be made arbitrarily close to $L$ by taking $x$ sufficiently close to $c$, and values of $g$ can be mare arbitrarily close to $M$ by taking $x$ sufficiently close to $c$, then values of $h$ can be made arbitrarily close to $L+M$ by taking $x$ sufficiently close to $c$.
What this means is: If $\lim_{x\to c} f(x) = L$ and $\lim_{x\to c} g(x) = M$, then $\lim_{x\to c} h(x) = L+M$. Now if we just replace $h$ with $f+g$, we get
$$
\lim_{x\to c}(f(x)+g(x)) = \lim_{x\to c} f(x) + \lim_{x\to c} g(x)
$$
In other words, the limit of a sum of functions is the sum of the limits of those functions, provided the limits of the individual functions exist.
Similarly, we can create new functions from old by subtraction, multiplication, division, and exponentiation. The limit laws show that (with a few exceptions, like you can't divide by zero), limits play nicely with algebra.
Why would anyone want to add together, divide, and square functions and limits though?
Fair question. Many, many functions can be expressed in terms of a few basic functions and these limit laws. For instance, think about these two basic limit statements.
\begin{align*}
\lim_{x\to c} x &= c \\
\lim_{x\to c} 1 &= 1
\end{align*}
Now what is $\lim_{x\to 3}(x^2-7)$? Using the limit laws, we can say:
\begin{align*}
\lim_{x\to3}(x^2-7)
&= \lim_{x\to 3}(x^2+(-7))
\\&= \lim_{x\to 3}(x^2) + \lim_{x\to 3}(-7)
\\&= \lim_{x\to 3}(x\cdot x) + \lim_{x\to 3}(-7)\cdot 1
\\&= \left(\lim_{x\to 3} x \right)\left(\lim_{x\to 3} x \right) + (-7)\lim_{x\to 3} 1
\\&= 3 \cdot 3 + (-7)\cdot 1 = 2
\end{align*}
You see how we didn't need anything other than the basic facts about the limits of $x$ and $1$, and the limit laws from the screenshot. And it feels right that the final limit is $(3)^2 - 7$; that is, the value of $f(x) = x^2-7$ at $x=3$. As we sometimes say, “plug in $3$” and you have the limit.
The functions that can be built from $x$, $1$, addition, scaling (multiplying by constants), and multiplication of functions is the set of polynomials. Using arguments generalizing the one above, we can now say that
$$
\lim_{x\to c}f(x) = f(c)
$$
when $f$ is any polynomial function. Cool trick!
We can throw in division with a slight caveat. A quotient of polynomials is called a rational function. We have to exclude from the domain of such a function any point which makes the denominator polynomial zero. But otherwise, we can use the limit laws again, and say $\lim_{x\to c}f(x) = f(c)$ when $f$ is any rational function and $c$ is in the domain of $f$.
This is a bit of a stretch, but one can argue that the rest of calculus is about limits of functions that can't be determined so easily with these limit laws.
Best Answer
The condition of continuity of $u$ is necessary as the following shows:
Define $ v(x)=x^2 $ for $ x\neq 0$, and $u(x)=1$ if $ x>0$, $u(x)=0$ if $ x\leq 0$. As you observe $\displaystyle\lim_{x\rightarrow0}u\circ v(x)=1$ while $ u(\displaystyle\lim_{x\rightarrow0}v(x))=0 $.
A condition for the equality $\displaystyle\lim_{x\rightarrow c}u\circ v(x)=\displaystyle\lim_{x\rightarrow L}u(x)$ when $\displaystyle\lim_{x\rightarrow c}v(x)=L$ (and $\displaystyle\lim_{x\rightarrow L}u(x)$ exists) is that either $ u $ is continuous at $L$ or there is a $ \delta >0$ s.t. $ \forall x(0<|x-c|<\delta\Rightarrow v(x)\neq L) $.