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$$\displaystyle \lim_{x \to \infty}\dfrac{8-\sqrt{x}}{8+\sqrt{x}}$$

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?

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  • $\begingroup$ Do you know L'Hopital's rule? $\endgroup$ – MCT Sep 9 '14 at 5:33
  • $\begingroup$ You have some typos(?) in your work. $\endgroup$ – Jonas Meyer Sep 9 '14 at 5:34
  • $\begingroup$ Factor out x from both and cancel the terms which has x in their denominator and so you find the answer to be - 1 $\endgroup$ – Avrham Aton Sep 9 '14 at 5:35
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    $\begingroup$ E.g., do you mean denominator? Do you mean to multiply with $\frac{8-\sqrt x}{8 - \sqrt x}$? Otherwise it doesn't make sense. $\endgroup$ – Jonas Meyer Sep 9 '14 at 5:44
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    $\begingroup$ @user437158: You can, but you did one and said the other. $\endgroup$ – Jonas Meyer Sep 9 '14 at 5:51
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Try dividing each term by $\sqrt{x}$ instead. Intuitively, it is the dominating term as $x$ gets large: $$ \lim_{x \to \infty}\dfrac{8-\sqrt{x}}{8+\sqrt{x}} = \lim_{x \to \infty}\dfrac{\frac{8}{\sqrt x}-1}{\frac{8}{\sqrt x}+1} = \dfrac{0-1}{0+1} = -1 $$

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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 $-1$.

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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 $$\require{cancel}\lim_{x\to\infty}\frac{8−\sqrt{x}}{8+\sqrt{x}}=\lim_{x\to\infty}\frac {-\cancel{\sqrt{x}}}{+\cancel{\sqrt{x}}}=−1$$

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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

  1. $\infty$ if $m>n$ and $\frac{a_m}{b_n}>0$;
  2. $-\infty$ if $m>n$ and $\frac{a_m}{b_n}<0$;
  3. $0$ if $n>m$
  4. $\dfrac{a_m}{b_n}$ if $m=n$.

In your case $m=n=1$, $a_1=-1$ and $b_1=1$.

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I'll add two more ways of solving.

  1. 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

    $$ \lim_{x \to \infty} \frac{8-x^{1/2}}{8+x^{1/2}} = \lim_{x \to \infty} \frac{-\frac 1 2 x^{-1/2}}{\frac 1 2 x^{-1/2}} = \lim_{x \to \infty} (-1) = -1 $$

  2. Let $t^2=x$. Then, $\lim_{x \to \infty} t = \lim_{x \to \infty} \sqrt{x} = \infty $. Consequently:

    $$ \lim_{x \to \infty} \frac{8-\sqrt x}{8+\sqrt x} = \lim_{t \to \infty} \frac{8-\sqrt {t^2}}{8+\sqrt {t^2}} = \lim_{t \to \infty} \frac{8-|t|}{8+|t|} = \lim_{t \to \infty} \frac{8-t}{8+t} = -1 $$

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