<s>[INST] Values of the limit $\lim\limits_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$ Hi I have a question regarding finding the values of limit for the following question.

Let $a, b \in \mathbb R$. Find the limit
$$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$

 [/INST] For a more informal answer...
$$\begin{align*}\sqrt{(x+a)(x+b)}-x&= \sqrt{x^2 + (a+b)x + ab} - x \\&= \sqrt{\left(\dfrac{a+b}{2} + x\right)^2-\frac{(a+b)^2}{4} +a b} - x \end{align*}$$
The expression $-\frac{1}{4} (-a-b)^2+a b$ is constant and is not important as $x \to \infty$ because the expression under the square root is dominated by the squared term.
Therefore
$$\begin{align*}\lim_{x\to\infty}\left(\sqrt{(x+a)(x+b)}-x\right)&= \\ \lim_{x\to\infty} \left(x + \dfrac{a+b}{2}\right) - x &= \dfrac{a+b}{2} \end{align*}$$
To add justification, consider $$\sqrt{ax^n + bx^m} = (ax^n + bx^m)^{1/2}, \quad n \geq m.$$
This is equal to $$(ax^n)^{1/2} + \frac{1}{2}\dfrac{bx^m}{\sqrt{ax^n}} - \dfrac{1}{8}\dfrac{b^2x^{2m}}{(\sqrt{ax^n})^3} ... $$
If we consider the case $n = 2, m = 1$ it is clear that the first two terms are non zero as $x \to \infty$ which is why we can't say that $\sqrt{x^2 - x} - x \approx 0 $ for large $x$.  On the other hand, the only term left in the case where $n = 2$ and $m = 0$ as $x \to \infty$ is the first term because every other term has a power of $x$ in the denominator.
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