Why can we modify expressions to use limits? If I wanted to take the limit of $\frac { \left( x\cdot \cos { \left( x \right) +\sin { \left( x \right)  }  }  \right)  }{ x+{ x }^{ 2 } } $ as x approaches 0,  I cannot do it directly as that would result in dividing by zero. However, if I modify the expression through several steps, I can take the limit of $$\frac { \left( \cos { \left( x \right) +\frac { \sin { \left( x \right)  }  }{ x }  }  \right)  }{ 1+{ x } } $$ as x approaches zero and get $2$. 
How can equal expressions give different results? 
And also: Why does $\\ \\ \lim _{ x\rightarrow 0 }{ \frac { \sin { \left( x \right)  }  }{ x } =1 } \\ $? Am I right in thinking that an infinitely small numerator cancels out an infinitely small denominator? 
 A: $\require{cancel}$ 
Because you should always remember that limit at $x=a$ means evaluation of function in a very small neighborhood of $x=a$ not exactly at $x=a$  
a limit is the value that a function or sequence "approaches" as the input or index approaches some value. 
Take into account the following function:
 
as $x\to 3^+$ meaning that we approach $3$ from right or from $+\infty$, the function value approaches $4$ so $\lim_{x\to 3^+}f(x) = 4$
but exactly at $x=3$ the value of the function is $1$ so $f(3)=1$
as $x\to 3^-$ meaning that we approach $3$ from left or from $-\infty$, the function value approaches $1$ so $\lim_{x\to 3^-}f(x) = 1$ 
Another example: consider the function $y=2\frac{x^2}{x}+3$ as you know the domain of the function is $\mathbb R-\{0\}$

as we approach $x=3$ from $+\infty$ the value of $y$ approaches $3$ so $\lim_{x\to 0^+}f(x)=3$ and as we approach $x=3$ from $-\infty$ the value of $y$ approaches $3$ so $\lim_{x\to 0^-}f(x)=3$ now
$$\text{because}\; \lim_{x\to 0^+}f(x)=\lim_{x\to 0^-}f(x)=3\Rightarrow \lim_{x\to 0}f(x)=3$$ so in spite of the fact that $f(x)$ is not definable at $x=0$ but the limit near $x=0$ is definable
consider $f(x)=\frac{\sin x}{x}$ again the domain is $\mathbb R-\{0\}$

but from the above diagram we see that as we approach $x=0$ from right the value of $f(x)$ approaches $1$ so $\lim_{x\to 0^+}f(x)=1$ and as we approach $x=0$ from left the value of $f(x)$ approaches $1$ so $\lim_{x\to 0^-}f(x)=1$ now $$\lim_{x\to 0^+}f(x)=\lim_{x\to 0^-}f(x)=1\Rightarrow \lim_{x\to 0}f(x)=1$$
again we cannot define $f(0)$ but we can define $\lim_{x\to 0}f(x)$
Now consider the following diagram:

The blue diagram is for $f(x)=\sin x$ and the red one is for $g(x)=x$
As you see when $x$ is infinitely near to $0$ not exactly at $0$ (the concept of limit) $f(x)\simeq g(x)$
meaning that at an infinitely small neighborhood of $x=0$ we have $\sin x\simeq x\Rightarrow \frac{sin x}{x}\simeq 1$ so
$$\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{x}{x}=1$$
in $\frac{x}{x}$ because $x$ is infinitely near to $0$ and so $x\neq 0$, we can cancel out $x$ fom the numerator and denumerator and say $\frac{x}{x}=\frac{\cancel{x}}{\cancel{x}}=1$ meaning that $\frac{x}{x}=1$ at $x\neq 0$ but infinitely near to $0$


*

*The concept of $\lim_{x\to a}f(x)$ is $x\neq a$. When you see
$\lim_{x\to a}f(x)$ you should understand that $x\neq a$ but
infinitely near to $a$  

*$f(x)=\sin x$ and $g(x)=x$ are different functions but in an infinitely small neighborhood of $x=0$ we have $f(x)\simeq g(x)$
meaning that they have the same limit at $x=0$ or $\lim_{x\to
   0}f(x)=\lim_{x\to 0}g(x)$ in limit calculation we can modify
expressions by using different function that have the same limit at a
special $x$
A: Perhaps looking at a simpler example might make things clearer for you.
What is the limit of $f(x) = 2$ as $x \to 0$? It's, well, 2, since $f(x)$ is, in fact, independent of $x$ (i.e., a constant).
Now consider the function $$g(x) = \frac{2x}{x}$$It's a different function from $f(x) = 2$ because, as written, it's not defined for $x=0$. However, so long as $x\ne0$, you can simplify $g(x)$ and write $g(x) = f(x)$, that is, both functions have the same value (namely, 2) for $x \ne 0$.
Since, when you take the limit $x \to 0$, you're not actually setting $x = 0$, the value of $g(x)$ will continue to be equal to that of $f(x)$, that is, 2, all the way to the limit (without ever being computed at $x=0$).
The point is, the functions $g(x) = 2x/x$ and $f(x) = 2$, albeit different, have the same limit as $x\to 0$.
It's exactly the same with the function that you have and the transformation that you applied. You found a different function that has the same limit as the function you started with, and you know that because outside of the point where you're computing the limit you have shown them to be identical functions, and your transformed function has a well-defined limit.
A: I think students are so used to "plugging the value of $x$" to calculate limits that they forget the fundamental aspect that a limit of a function is not necessarily a value of the function but it is related to the values of a function.
The definition of limit does not give any method to evaluate limits, rather it gives us a mechanism to check (at least theoretically) whether a given number is (or is not) the limit of a function. However the definition is sufficient to evaluate the following simple limits: $$\lim_{x \to a}k = k,\,\lim_{x \to a}x = a,\,\lim_{x \to \infty}\frac{1}{x} = 0\tag{1}$$ Then the usual route to evaluation of limits is not through plugging the value of $x$, but using algebra of limits.
The theorems which come under algebra of limits explicitly mention the conditions under which they can be used effectively. The most imposing limitations of these rules are the following:
1) limit of sub-expressions must exist (by that we mean that it is a finite number).
2) limit of an expression occurring in denominator is non-zero.
Normally the textbook / exam questions are designed in such manner that the rules of algebra of limits can't be used directly (because of the above limitations). Then we are given some more tools to help us in evaluating limits:
1) Squeeze theorem: If $g(x)\leq f(x)\leq h(x)$ in a deleted neighborhood of $a$ and $\lim_{x \to a}g(x) = \lim_{x \to a}h(x) = L$ then $\lim_{x \to a}f(x) = L$.
2) The Standard Limits:
$$\lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}\frac{e^{x} - 1}{x} = \lim_{x \to 0}\frac{\log(1 + x)}{x} = 1,\,\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{2}$$
3) The expression can be manipulated using algebraical/trigonometrical identities (or identities pertaining to specific functions involved in the expression) under the assumption that $x \neq a$ (when we are evaluating the limit as $x \to a$).
The goal here to transform the expression in such a form where we can make use of rules of "algebra of limits", the simple limits of equation $(1)$ and  "the standard limits" mentioned in equation $(2)$. Note that these transformations are based on identities so that new expression is equal to the old expression (just like $1 = 2/2$). Also note that when we evaluate a limit as $x \to a$ we are totally ignorant of what happens when $x = a$ so all the transformations are performed with assumption that $x \neq a$.

In the current question if we apply the "rules of algebra of limits" we face the limitation of zero denominator. Hence we need to "transform the expression" in a manner so that the zero denominator is avoided and it is possible to make use of the standard limits. This is what you have done when you have divided numerator and denominator of the given expression by $x$.
This clearly gets rid of zero denominator and also gives the expression $(\sin x)/x$ so that we can make use of standard limits. However there is also the term $\cos x$ in numerator. The limit of this term is not evaluated by plugging $x = 0$ but rather it is evaluated by using the following result $$\lim_{x \to a}\cos x = \cos a\tag{3}$$ which can be proven using the standard limit $(\sin x)/x \to 1$ as $x \to 0$. So there are some nice functions like $\cos x$ for which the limit can be really evaluated by plugging. Such functions are called continuous functions and most of the usual functions in calculus are continuous at points where they are defined. This justifies the "plugging".
However a student learning calculus must understand very clearly that plugging does not work by default in evaluation of limits, but rather depends on certain functions being continuous. And one must understand the proof that usual functions in calculus are continuous (these proofs are based on standard limits in $(2)$).

BTW your statement about "infinitely small" is "nothing". You need to understand that there is no real number which is infinitely small (or infinitely large). These terms are coming from an era when there was no sound theory of calculus. At least in the current century we have a sound theory of calculus and it is better to treat such phrases as non-sensical and not waste time on them.
A: Don't be concerned with the value of the function at $x = 0.$ Unless the function is continuous, it has little or nothing to do with the value of the limit of the function as $x$ approaches $0$.
If $\lim f(x)$ and $\lim g(x)$ exists, then
$\lim f(x) g(x) = \lim f(x) \lim g(x)$
$\lim \dfrac{f(x)}{g(x)} = \dfrac{\lim f(x)}{\lim g(x)}$ 
(Provided $\lim g(x) \ne 0$)
$$
\lim_{x \to 0} \dfrac{x \cos x + \sin x}{x+ x^2}
= \left( \lim_{x \to 0} \dfrac xx \right)\;
   \left(\lim_{x \to 0} \dfrac{\cos x +\frac{\sin x}{x}}{1+x}\right)
= (1) \left(\dfrac{1+1}{1+0}\right) = 2
$$
