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.

$$(\sqrt{x^2 + 3x} - \sqrt{x^2 + 1}) =(x \sqrt{1 + 3/x} - x\sqrt{1+ x^{-2} })$$
$$
=x\left( \sqrt{1 + 3/x} - \sqrt{1+ x^{-2} }\right)
$$
$$
\left( \sqrt{1 + 3/x} - \sqrt{1+ x^{-2} }\right) =1 +\frac{3}{2}\frac{1}{x} + O(x^{-2}) - (1+ O(x^{-2}))
$$
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. 
