Q:

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:

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:

$\require{cancel}$ 
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 $x$

