Calculate $\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$ without using L'Hospital's rule Question:
Calculate
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$$
without using L'Hospital's rule.
Attempted solution:
First we multiply with the conjugate expression:
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1}) = \lim_{x \to \infty} \frac{x^2 + 3x - (x^2 + 1)}{\sqrt{x^2 + 3x} + \sqrt{x^2 + 1}}$$
Simplifying gives:
$$\lim_{x \to \infty} \frac{3x - 1}{\sqrt{x^2 + 3x} + \sqrt{x^2 + 1}}$$
Breaking out $\sqrt{x^2}$ from the denominator and $x$ from the numerator gives:
$$\lim_{x \to \infty} \frac{x(3 - \frac{1}{x})}{x(\sqrt{1 + 3x} + \sqrt{1 + 1})} = \lim_{x \to \infty} \frac{3 - \frac{1}{x}}{\sqrt{1 + 3x} + \sqrt{2}}$$
The result turns out to be $\frac{3}{2}$, but unsure how to proceed from here.
 A: There's an error in the factorisation of the denominator. No need to factor $x$ in the numerator, as you have a theorem for the limit at $\infty$ of a rational function. You should obtain, if $x>0$:
$$\lim_{x \to+ \infty}\frac{3x-1}{x\Bigl(\sqrt{1 + \dfrac3x} + \sqrt{1 + \dfrac1{x^2}}\Bigr)} = \lim_{x \to +\infty} \frac{3x - 1}x\cdot\frac1{\sqrt{1 + \dfrac3x} + \sqrt{1 + \dfrac1{x^2}}}= 3\cdot\frac12.$$
Similarly the limit as $x\to-\infty\,$ is $\,-\dfrac32$.
A: Your extraction of $x$ is incorrect: what you need is
$$ \sqrt{x^2+3x} = \sqrt{x^2(1+3/x)} = x\sqrt{1+3/x}. $$
Then you have
$$ \lim_{x \to \infty} \frac{3x-1}{x(\sqrt{1+3/x}+\sqrt{1+1/x^2})} = \lim_{x \to \infty} \frac{3}{\sqrt{1+3/x}+\sqrt{1+1/x^2}} - \frac{1}{x(\sqrt{1+3/x}+\sqrt{1+1/x^2})} $$
The square roots in this expression all tend to $1$, so the first term tends to $3/2$, the second to $0$.
A: I do not agree with your result after factorizing in the last step. I get 
$$\frac{x(3-1/x)}{x\left(\sqrt{1+\frac{3}{x}}+\sqrt{1+\frac{1}{x^2}}\right)}=\frac{3-1/x}{\sqrt{1+\frac{3}{x}}+\sqrt{1+\frac{1}{x^2}}}\rightarrow 3/2$$
A: $$
\sqrt{x^2+3x} - \sqrt{x^2+1} = x\sqrt{1+\frac{3}{x}} - x\sqrt{1+\frac{1}{x^2}} = x\left(\sqrt{1+\frac{3}{x}} -\sqrt{1+\frac{1}{x^2}} \right)
$$
expand the radicals we find
$$
\sqrt{1+\frac{3}{x}} = 1 + \frac{3}{2x} + O(x^{-2})\\
\sqrt{1+\frac{1}{x^2}}  = 1 + O(x^{-2})
$$
put it all together
$$
\lim_{x\to\infty}\left(\sqrt{x^2+3x} - \sqrt{x^2+1}\right) \to x\left(1 + \frac{3}{2x}-1\right) = \frac{3}{2}
$$
A: $$(\sqrt{x^2 + 3x} - \sqrt{x^2 + 1}) =(x \sqrt{1 + 3/x} - x\sqrt{1+ x^{-2} })$$
$$
=x\left( \sqrt{1 + 3/x} - \sqrt{1+ x^{-2} }\right)
$$
$$
\left( \sqrt{1 + 3/x} - \sqrt{1+ x^{-2} }\right) =1 +\frac{3}{2}\frac{1}{x} + O(x^{-2}) - (1+ O(x^{-2}))
$$
using Taylor's theorem. 
So we get 
$$
\frac{3}{2} + O(x^{-1})
$$
which converges to $3/2.$ 
Note that this approach is more easily generalizable to cases of different powers, eg cube roots. 
