Limit of $\frac{x}{a+x}$ Why is it that when I look at $$\lim_{x\to\infty} \frac{x}{a+x}$$ 
I don't know the answer because it's an  indeterminate form, but if I divide by x 
$$\lim_{x\to\infty} \frac{1}{\frac{a}{x}+1}$$
the answer appears "magically"  ?
What happened ?
 A: There is no mystery in the phrase “indeterminate form”: it just means that you have a limit where you can't directly apply any of the general theorems on limits that can be instead applied in other situations.
For instance,
$$
\lim_{x\to1}\frac{2x+1}{3x+2}=\frac{3}{5}
$$
because you know the theorem that says

if both $\lim_{x\to a}f(x)=l$ and $\lim_{x\to a}g(x)=m$ exist, are finite and $m\ne0$, then $$\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{l}{m}$$

(Note that $a$ can be $\infty$ or $-\infty$.)
This theorem can be supplemented with other cases; for instance, if $l=\infty$ and $m>0$ (but still finite), then $\lim_{x\to a}\frac{f(x)}{g(x)}=\infty$. Other criteria can be applied when $m=0$ (with some care).
However, no general criterion exists when both limits are $0$ or both are infinity ($\infty$ or $-\infty$).
In these cases one can get away by rewriting the function in a different form and this is exactly the “magic” you did.
Since you're considering the limit at $\infty$, you can assume the function is defined in some interval $(b,\infty)$ so that $b>0$ and $b>-a$. Hence, for $x\in(b,\infty)$, we can write
$$
\frac{x}{a+x}=\frac{1}{1+\frac{a}{x}}
$$
and now we can apply the above theorem, because the numerator and the denominator both have limit $1$.
We could apply l'Hôpital, but it would be overkill:
$$
\lim_{x\to\infty}\frac{x}{a+x}=\lim_{x\to\infty}\frac{1}{1}=1
$$
A: Hint:
The result is not ''magic'' but a consequence of the property:
$$
\lim_{x\rightarrow p}\dfrac{f(x)}{g(x)}=\dfrac{\lim_{x\rightarrow p}f(x)}{\lim_{x\rightarrow p}g(x)}
$$
Use this (two fold) to evalate
$$
\lim_{x\rightarrow +\infty}\dfrac{\frac {f(x)}{x}}{\frac {g(x)}{x}}
$$
A: The value of $\frac{x}{a+x}$ at $x=\infty$ is indeterminate. 
But 
The value of $\lim_{x\to\infty} \frac{x}{a+x}=1$
Or 
The value of $\lim_{x\to\infty} \frac{1}{\frac{a}{x}+1}=1$
Above two equations are equivalent. 
It's just that if we put directly $x=\infty$ in the given expression then we won't get any value because no value of $\frac{x}{a+x}$ exists at $x=\infty$ but if we want the closest value of $$\frac{x}{a+x}$$ as $x$ approaches $\infty$ then we'll get $1$.
It's the essence of limit and follows from definition of limit. 
