How x times a square root grows at infinity
Analyze the behavior of the function $$ y = x\sqrt{\frac{x-1}{x+1}} $$ as $x \to +\infty$.
Explanation of Method
Dividing the numerator and denominator inside the square root by $x$ shows that the fraction under the root tends to $1$, so the square root factor itself tends to $1$ while contributing negligibly little extra information about the rate of growth of $y$. To pin down not just that $y\to\infty$ but exactly how, we push the approximation one order further and find a linear asymptote for $y$.
- Simplify the fraction $\dfrac{x-1}{x+1}$ by dividing through by $x$.
- Take the limit of the square root as $x \to +\infty$ to see that $y \sim x$.
- Refine the approximation to find the oblique asymptote of $y$.
Step 1 · Simplify the fraction under the root
Dividing numerator and denominator by $x$:
$$ \frac{x-1}{x+1} = \frac{1 - \frac{1}{x}}{1 + \frac{1}{x}}. $$
As $x \to +\infty$, $\dfrac{1}{x} \to 0$, so:
$$ \frac{1-\frac{1}{x}}{1+\frac{1}{x}} \to \frac{1-0}{1+0} = 1, \qquad \text{hence} \qquad \sqrt{\frac{x-1}{x+1}} \to \sqrt{1} = 1. $$
Step 2 · Leading-order growth of $y$
Since the square root factor tends to $1$, for large $x$:
$$ y = x\sqrt{\frac{x-1}{x+1}} \approx x \cdot 1 = x. $$
Because $x \to +\infty$ and the square-root factor stays bounded near $1$, the product $y$ also grows without bound:
$$ \lim_{x \to +\infty} x\sqrt{\frac{x-1}{x+1}} = +\infty. $$
Step 3 · Refine the approximation (oblique asymptote)
To see precisely how $y$ compares to $x$, rewrite $\dfrac{x-1}{x+1} = 1 - \dfrac{2}{x+1}$ and apply the binomial approximation $(1-u)^{1/2}\approx 1-\tfrac{u}{2}$ for small $u = \dfrac{2}{x+1}$:
$$ \sqrt{\frac{x-1}{x+1}} = \sqrt{1 - \frac{2}{x+1}} \approx 1 - \frac{1}{x+1}. $$
Multiplying by $x$:
$$ y \approx x\left(1 - \frac{1}{x+1}\right) = x - \frac{x}{x+1}. $$
As $x \to +\infty$, $\dfrac{x}{x+1} \to 1$, so:
$$ y - x \to -1, \qquad \text{i.e.} \qquad y \approx x - 1 \ \text{ for large } x. $$
Summary Table
| Quantity | Behavior as $x \to +\infty$ |
|---|---|
| $\dfrac{x-1}{x+1}$ | $\to 1$ |
| $\sqrt{\dfrac{x-1}{x+1}}$ | $\to 1$, more precisely $\approx 1-\dfrac{1}{x+1}$ |
| $y = x\sqrt{\dfrac{x-1}{x+1}}$ | $\to +\infty$, with oblique asymptote $y \approx x-1$ |
Additional Notes
- Saying $\sqrt{\frac{x-1}{x+1}} \to 1$ only shows that the extra factor is asymptotically negligible in a multiplicative sense; it does not immediately reveal how $y$ compares to $x$ in an additive sense — that requires the finer expansion in Step 3.
- The result $y \approx x - 1$ means the graph of $y = x\sqrt{\frac{x-1}{x+1}}$ has the line $y = x-1$ as an oblique (slant) asymptote as $x \to +\infty$.
- The domain of $y$ requires $\dfrac{x-1}{x+1} \ge 0$, which holds for $x \ge 1$ or $x < -1$; here we only consider the branch relevant to $x \to +\infty$.
Result
As $x \to +\infty$, the function grows without bound: