Values of the limit $\lim\limits_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$

Question

Hi I have a question regarding finding the values of limit for the following question.

Let $a, b \in \mathbb R$. Find the limit

$$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$

Answer 1

Hint: Multiply the numerator and denominator by the conjugate of the expression.

That is, you can try working with $$\dfrac{\sqrt{(x+a)(x+b)} - x}1\cdot\frac{\sqrt{(x + a)(x + b)} + x}{\sqrt{(x + a)(x + b)} + x} = \dfrac{(x+a)(x + b) - x^2}{\sqrt{(x + a)(x+b)} + x}$$

Expanding $(x + a)(x + b)$ in the numerator and simplifying gives us $$ \dfrac{(a+b)x + ab}{\sqrt{x^2 + (a+b)x + ab} + x}$$

Now divide numerator and denominator by $x$ to find your limit.

$$ \lim_{x\to \infty} \dfrac{(a+b) + \frac{ab}x}{\sqrt{1 + \frac{(a+b)}x + \frac{ab}{x^2}} + 1} = \dfrac{a+b}{2}$$

Answer 2

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.