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: $$\lim_{x\to \infty}\frac{9x}{\sqrt{x^2 + 4x} + \sqrt{x^2-5x}}$$
$$\\ \lim _{ x\rightarrow \infty }{ \sqrt { { x }^{ 2 }+4x } -\sqrt { { x }^{ 2 }-5x } } =\lim _{ x\rightarrow \infty }{ \frac { \left( \sqrt { { x }^{ 2 }+4x } -\sqrt { { x }^{ 2 }-5x } \right) \left( \sqrt { { x }^{ 2 }+4x } +\sqrt { { x }^{ 2 }-5x } \right) }{ \sqrt { { x }^{ 2 }+4x } +\sqrt { { x }^{ 2 }-5x } } } =\\ =\lim _{ x\rightarrow \infty }{ \frac { 9x }{ \sqrt { { x }^{ 2 }+4x } +\sqrt { { x }^{ 2 }-5x } } } =\lim _{ x\rightarrow \infty }{ \frac { 9x }{ \sqrt { { x }^{ 2 }\left( 1+\frac { 4 }{ x } \right) } +\sqrt { { x }^{ 2 }\left( 1-\frac { 5 }{ x } \right) } } } =\lim _{ x\rightarrow \infty }{ \frac { 9x }{ \left| x \right| \left( \sqrt { 1+\frac { 4 }{ x } } +\sqrt { 1-\frac { 5 }{ x } } \right) } } =\\ \overset { if\quad x>0 }{ = } \lim _{ x\rightarrow \infty }{ \frac { 9x }{ x\left( \sqrt { 1+\frac { 4 }{ x } } +\sqrt { 1-\frac { 5 }{ x } } \right) } } =\lim _{ x\rightarrow \infty }{ \frac { 9 }{ \left( \sqrt { 1+\frac { 4 }{ x } } +\sqrt { 1-\frac { 5 }{ x } } \right) } } =\frac { 9 }{ \sqrt { 1+0 } +\sqrt { 1-0 } } =\frac { 9 }{ 2 } $$
$$\lim_{x \to \infty} \sqrt{(x^2)+4x} - \sqrt{(x^2)-5x}$$
$$\lim_{x \to \infty} {(x^2)+4x - (x^2)+5x\over\sqrt{(x^2)+4x} +\sqrt{(x^2)-5x}}$$
$$\lim_{x \to \infty} {9x\over\sqrt{(x^2)+4x} +\sqrt{(x^2)-5x}}$$
$$\lim_{x \to \infty} {9\over\sqrt{(x^2)/x^2+4x/x^2} +\sqrt{(x^2)/x^2-5x/x^2}}$$
$$\lim_{x \to \infty} {9\over\sqrt{1+4/x} +\sqrt{1-5/x}}$$
$${9\over\sqrt{1} +\sqrt{1}}={9\over 2}$$
Multiply and divide by $$\sqrt{x^2+4x} +\sqrt{x^2-5x}$$ $$\lim_{x\to \infty} \frac{9x}{\sqrt{x^2+4x} +\sqrt{x^2-5x}}$$ $$\lim_{x\to \infty} \frac{9}{\frac{\sqrt{x^2+4x}}{x} +\frac{\sqrt{x^2-5x}}{x}}$$
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 $
<p>$\lim_{x\rightarrow{c}} g(x) = 0$</p> <p>$\lim_{x\rightarrow{c}}\frac{f'(x)}{g'(x)}=L$</p>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.
This is a difference $f(y)-f(z)$ where the difference $y-z$ compared to $y$ and $z$ is small. It sounds like a derivative, so can we try something to make it one? What if we use substitution $h=1/x$?
$$ \begin{align} &\lim_{x \to \infty} \sqrt{x^2 + 4x} - \sqrt{x^2 - 5x} = \lim_{x \to \infty} x \left( \sqrt{1 + 4/x} - \sqrt{1 - 5/x} \right) \\ =& \lim_{h \to 0^+} \frac{\sqrt{1 + 4h} - \sqrt{1 - 5h}}{h} \end{align} $$ Now it's almost like the derivative of $g(x) = \sqrt{x}$ at $1$, only that we have different coefficients $1+4h$ and $1-5h$ instead of $1+h$ and $1$ there.
The value of this kind of limit limit is actually $(+4) - (-5)$ times the derivative of $\sqrt{x}$ at $1$. (I'll show below why.) The derivative is $g'(x) = 1/(2\sqrt{x})$, so the answer is $$ 9 g'(1) = \frac{9}{2} $$
Here's why you can do this: We will see the two limit exist so splitting the first limit is legal, and then we make a change of variables $s=\alpha h$ and $t=-\beta h$: $$ \begin{align} &\lim_{h \to 0} \frac{g(x_0+\alpha h) - g(x_0+\beta h)}{h} \\ =&\lim_{h \to 0} \frac{g(x_0+\alpha h) - g(1) + g(1) - g(x_0+\beta h)}{h} \\ =& \lim_{h \to 0} \frac{g(x_0+\alpha h) - g(x_0)}{h} - \lim_{h \to 0} \frac{g(x_0 + \beta h) - g(x_0)}{h} \\= & \lim_{s \to 0} \alpha\frac{g(x_0+t) - g(x_0)}{s} - \lim_{t \to 0} \beta\frac{g(x_0+t) - g(x_0)}{t} \\= & \alpha g'(x_0) - \beta g'(x_0) = (\alpha-\beta) g'(x_0) \end{align} $$
Addendum: As pointed out in the comments, a shorter way to show this is to define another function $\hat{g}(x) = g(x_0+\alpha x) - g(x_0+\beta x)$ and notice that $\hat{g}(0)=0$ so we have $$ \lim_{h \to 0} \frac{g(x_0+\alpha h) - g(x_0+\beta h)}{h} = \lim_{h \to 0} \frac{\hat{g}(h) - \hat{g}(0)}{h} = \hat{g}'(0) = \alpha g'(x_0) - \beta g'(x_0) $$
