<s>[INST] 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.
 [/INST] 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 $0$.
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