Avoiding the limit notation during long algebraic manipulations

Solution 1:

Having commented, edited and almost commented again, I realize it is better if I will write everything in an organized manner.

1. Manipulate the expression using only 'proper' equalities (equailities that don't involve limits) and only then take the limit. For example: $$ \frac{x+1}{x} = 1 + \frac{1}{x} \xrightarrow[x\rightarrow\infty]{} 1$$
2. If we were to look at your example, we could essentially rewrite it similarly to what we did in point 1, without writing $\text{lim}$ so many times: $$\frac{x}{x+1}=\frac{\frac{x}{x}}{\frac{x}{x}+\frac{1}{x}}\xrightarrow[x\rightarrow\infty]{}\frac{1}{1+0}={\frac{1}{1}}=1$$
3. Write $\text{lim}$ without the subscript '$x→∞$' (but then you have to remember that $x$ tends to $∞$).
4. Once you reach something of the form $\lim_{x\rightarrow \infty}\frac{1}{1+0}$ then it is widely accepted to just write it equates to $1$ (Without further explaining).
5. Sometimes it is of great convinience to use O-notation. Use of that little-O would look like: $$\frac{x+1}{x} = 1+ o(1)\quad(x\rightarrow\infty) $$ And big-O: $$ \frac{x+1}{x} =1+\mathcal{O}\left(\frac{1}{x}\right)\quad(x\rightarrow\infty) $$ Having brought the expression to a final form (which resembles the 2 latter equalities) one immediately recognizes the limit. This notation, usually, conceals some limiting procedure.

Of course, the above guidelines only suggest a way of having a cleaner, more compact paper, and are not of any mathematical importance. The basic rule is; the purpose of mathematical notations is to convey an ideas or objects. [I would say] As a rule of thumb: Notations are good if they are intuitive (and intuitevely understandable), neat and simple.

Solution 2:

I would write like this:

$$ \frac{x}{x+1} = {\frac{\frac{x}{x}}{\frac{x}{x}+\frac{1}{x}}} \hspace{1em} \xrightarrow[x\to \infty]{} \hspace{1em} 1 $$

Solution 3:

You could just use language:

We note that $\frac{x}{x+1}=\frac{\frac{x}{x}}{\frac{x}{x}+\frac{1}{x}} \to 1$ as $x \to \infty$, since $\frac{1}{x}$ vanishes in the denominator.

Solution 4:

Apart from the long arrow notation, mentioned by others: $$ f \xrightarrow[x\to\infty]{} g$$ I would recommend, in this specific problem, to use a new variable. Note that for $$u = x+1$$ we have $$\frac x{x+1} = \frac {u-1}u = 1-\frac 1u$$ hence $$ \lim_{x\to\infty} \frac x{x+1} = \lim_{u\to\infty} \left(1-\frac 1u\right) = 1-0 = 1$$