<s>[INST] How can I evaluate: $\lim_{x \to \infty} \sqrt{x^2 + 4x} - \sqrt{x^2 - 5x}$ I need to find this limit:
$$\lim_{x \to \infty} \sqrt{x^2 + 4x} - \sqrt{x^2 - 5x}$$
The answer I got from using the limit laws is $\sqrt{\infty} - \sqrt{\infty}$.
How do I proceed now?
Added I took the conjugate of the function and I got a new and probably better function to work with: $$\lim_{x\to \infty}\frac{9x}{\sqrt{x^2 + 4x} + \sqrt{x^2-5x}}$$
 [/INST] If you know Differential Calculus, you may also use L'hopital's Rule to solve for the limit. Just as a recap, one instance of L'hopital's Rule states:

Given $\frac{f(x)}{g(x)}$ with some arbitrary limit to $c$ imposed on it, if all of the following constraints are met:

$\lim_{x\rightarrow{c}} f(x) = 0 $ 
$\lim_{x\rightarrow{c}} g(x) = 0$
$\lim_{x\rightarrow{c}}\frac{f'(x)}{g'(x)}=L$

then

$\lim_{x\rightarrow{c}}\frac{f(x)}{g(x)} = L$


Now, we have
$$\lim_{x\rightarrow\infty}\sqrt{x^2+4x}-\sqrt{x^2-5x}$$
which is obviously not in the form of $\frac{f(x)}{g(x)}$ as needed. However, we can put it into one, like so, as long as $x>0$:
$$\lim_{x\rightarrow\infty}\sqrt{x^2+4x}-\sqrt{x^2-5x} = $$
$$\lim_{x\rightarrow\infty}x\Bigg(\sqrt{1+\frac{4}{x}}-\sqrt{1-\frac{5}{x}}\Bigg)=$$
$$\lim_{x\rightarrow\infty}\frac{\sqrt{1+\frac{4}{x}}-\sqrt{1-\frac{5}{x}}}{x^{-1}} = \frac{0}{0}$$
Now we have the form we want and we found the limit, which is in the indeterminite form $\frac{0}{0}$. We also have our two functions, which are 
$$f(x)=\sqrt{1+\frac{4}{x}}-\sqrt{1-\frac{5}{x}}$$
$$g(x)=x^{-1}=\frac{1}{x}$$
Now we have to do find the derivative of $f(x)$ and $g(x)$, which are:
$$f'(x)=\frac{d}{dx}\bigg(1+\frac{4}{x}\bigg)^{0.5}\frac{d}{dx}\bigg(1+\frac{4}{x}\bigg)-\frac{d}{dx}\bigg(1-\frac{5}{x}\bigg)^{0.5}\frac{d}{dx}\bigg(1-\frac{5}{x}\bigg)=$$
$$-\frac{4}{2x^{2}\sqrt{1+\frac{4}{x}}}-\frac{5}{2x^{2}\sqrt{1-\frac{5}{x}}}=$$
$$\frac{-4\sqrt{1-\frac{5}{x}}-5\sqrt{1+\frac{4}{x}}}{2x^{2}\sqrt{1+\frac{4}{x}}\sqrt{1-\frac{5}{x}}}$$
$$g'(x)=-x^{-2}=-\frac{1}{x^2}$$
And finally
$$\lim_{x\rightarrow\infty}\frac{f'(x)}{g'(x)}=\lim_{x\rightarrow\infty}\frac{\Bigg(\frac{-4\sqrt{1-\frac{5}{x}}-5\sqrt{1+\frac{4}{x}}}{2x^{2}\sqrt{1+\frac{4}{x}}\sqrt{1-\frac{5}{x}}}\Bigg)}{-\frac{1}{x^2}}=$$
$$\lim_{x\rightarrow\infty}x^{2}\Bigg(\frac{4\sqrt{1-\frac{5}{x}}+5\sqrt{1+\frac{4}{x}}}{2x^{2}\sqrt{1+\frac{4}{x}}\sqrt{1-\frac{5}{x}}}\Bigg)=$$
$$\lim_{x\rightarrow\infty}\frac{4\sqrt{1-\frac{5}{x}}+5\sqrt{1+\frac{4}{x}}}{2\sqrt{1+\frac{4}{x}}\sqrt{1-\frac{5}{x}}}=$$
$$\frac{4\sqrt{1}+5\sqrt{1}}{2\sqrt{1}\sqrt{1}}=\frac{9}{2}$$
We have our answer based on L'hopital's Rule. I know this was not required, as there is a much easier algebraic manner to solve for the limit, but I thought it was nice trying to find the limit this way. 
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