How to evaluate $\displaystyle \lim_{x \to \infty}\frac{8-\sqrt{x}}{8+\sqrt{x}}$ $$\displaystyle \lim_{x \to \infty}\dfrac{8-\sqrt{x}}{8+\sqrt{x}}$$ 
I tried rationalizing the numerator: 
$$\lim_{x \to \infty}\dfrac{8-\sqrt{x}}{8+\sqrt{x}} \times \dfrac{(8-\sqrt{x})}{(8-\sqrt{x})}$$  
$$\lim_{x \to \infty}\dfrac{64-16\sqrt{x}+x}{64-x}$$
Is this correct? how do I proceed from here? 
 A: We can rewrite the expression as $$\lim \limits_{x \to \infty} (-1 + \frac{16}{8 + \sqrt x})$$
The second term goes to zero, thus the limit is $-1$.
A: Another approach, although this may not be rigorous:
As $x\to \infty$, $\sqrt{x}$ dominates, being the highest power in both  the numerator and denominator, hence
$$\require{cancel}\lim_{x\to\infty}\frac{8−\sqrt{x}}{8+\sqrt{x}}=\lim_{x\to\infty}\frac {-\cancel{\sqrt{x}}}{+\cancel{\sqrt{x}}}=−1$$
A: Try dividing each term by $\sqrt{x}$ instead. Intuitively, it is the dominating term as $x$ gets large:
$$
\lim_{x \to \infty}\dfrac{8-\sqrt{x}}{8+\sqrt{x}}
= \lim_{x \to \infty}\dfrac{\frac{8}{\sqrt x}-1}{\frac{8}{\sqrt x}+1}
= \dfrac{0-1}{0+1}
= -1
$$
A: First of all get rid of the square root: since squaring is a continuous function and $\lim_{t\to\infty}t^2=\infty$, you have
$$
\lim_{x\to\infty}\frac{8-\sqrt{x}}{8+\sqrt{x}}=
\lim_{t\to\infty}\frac{8-\sqrt{t^2}}{8+\sqrt{t^2}}=
\lim_{t\to\infty}\frac{8-t}{8+t}
$$
Now you have reduced to the limit at infinity of a rational function, for which there's an easy criterion. Suppose $a_m\ne0$ and $b_n\ne0$; then
$$
\lim_{t\to\infty}
  \frac{a_mt^m+a_{m-1}t^{m-1}+\dots+a_0}{b_nt^n+b_{n-1}t^{n-1}+\dots+b_0}
$$
can be rewritten as
$$
\lim_{t\to\infty}
  \frac{t^m\left(a_m+\dfrac{a_{m-1}}{t}+\dots+\dfrac{a_0}{t^m}\right)}
       {t^n\left(b_n+\dfrac{b_{n-1}}{t}+\dots+\dfrac{b_0}{t^n}\right)}
=\lim_{t\to\infty}\frac{a_mt^m}{b_nt^n}
$$
So this limit is


*

*$\infty$ if $m>n$ and $\frac{a_m}{b_n}>0$;

*$-\infty$ if $m>n$ and $\frac{a_m}{b_n}<0$;

*$0$ if $n>m$

*$\dfrac{a_m}{b_n}$ if $m=n$.


In your case $m=n=1$, $a_1=-1$ and $b_1=1$.
A: I'll add two more ways of solving.


*

*L'Hôpital's rule states that if $\lim_{x \to \infty}{f(x)} = \lim_{x
    \to \infty}{g(x)} = \infty$ then
$$ \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty}
    \frac{f'(x)}{g'(x)} $$
In this case
$$
    \lim_{x \to \infty} \frac{8-x^{1/2}}{8+x^{1/2}} = \lim_{x \to \infty}
    \frac{-\frac 1 2 x^{-1/2}}{\frac 1 2 x^{-1/2}} = \lim_{x \to \infty} (-1) = -1
    $$

*Let $t^2=x$. Then, $\lim_{x \to \infty} t = \lim_{x \to \infty} \sqrt{x} = \infty $. Consequently:
$$
    \lim_{x \to \infty} \frac{8-\sqrt x}{8+\sqrt x} = \lim_{t \to \infty} \frac{8-\sqrt {t^2}}{8+\sqrt {t^2}} = \lim_{t \to \infty} \frac{8-|t|}{8+|t|} = \lim_{t \to \infty} \frac{8-t}{8+t} = -1
    $$
