Limit of nested square roots minus sqrt(x)
Evaluate the limit: $$ L = \lim_{x \to \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}} - \sqrt{x}\right). $$
Explanation of Method
As $x\to\infty$, both $\sqrt{x+\sqrt{x+\sqrt x}}$ and $\sqrt x$ tend to $+\infty$, giving the indeterminate form $\infty-\infty$. Rationalizing (multiplying and dividing by the conjugate sum) is the standard way to turn a difference of two square roots into a fraction, which here collapses the outer layer of nesting and leaves a single inner expression, $\sqrt{x+\sqrt x}$, in the numerator. Dividing every remaining square root by $\sqrt x$ then converts each piece into a quantity with a clean, easily evaluated limit, without needing any further approximation.
- Rationalize the original difference using the conjugate $\sqrt{x+\sqrt{x+\sqrt x}}+\sqrt x$.
- Simplify the resulting numerator using the difference of squares.
- Divide numerator and denominator by $\sqrt x$ and evaluate each piece's limit.
- Combine the results to find $L$.
Step 1 · Rationalize the original expression
Let $A = \sqrt{x+\sqrt{x+\sqrt x}}$ and $B=\sqrt{x+\sqrt x}$, so that $A=\sqrt{x+B}$ and $A^2 = x+B$. Multiplying $L=A-\sqrt x$ by the conjugate $\dfrac{A+\sqrt x}{A+\sqrt x}$:
$$ A - \sqrt x = \frac{A^2-x}{A+\sqrt x} = \frac{(x+B)-x}{A+\sqrt x} = \frac{B}{A+\sqrt x}. $$
Step 2 · The expression is now a single ratio
So:
$$ L = \lim_{x\to\infty} \frac{B}{A+\sqrt x} = \lim_{x\to\infty} \frac{\sqrt{x+\sqrt x}}{\sqrt{x+\sqrt{x+\sqrt x}}+\sqrt x}. $$
Both the numerator and denominator still grow like $\sqrt x$, giving another indeterminate form $\dfrac{\infty}{\infty}$ — but now there is no more nesting to unravel, only ordinary square roots of $x$.
Step 3 · Divide everything by $\sqrt x$
Divide the numerator and every term of the denominator by $\sqrt x$ (valid for $x>0$):
$$ L = \lim_{x\to\infty} \frac{\dfrac{\sqrt{x+\sqrt x}}{\sqrt x}}{\dfrac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt x}+1} = \lim_{x\to\infty} \frac{\sqrt{1+\dfrac{1}{\sqrt x}}}{\sqrt{1+\dfrac{B}{x}}+1}, $$
where each square root has been pulled inside using $\dfrac{\sqrt{x+c}}{\sqrt x}=\sqrt{1+\dfrac cx}$ (with $c=\sqrt x$ for the numerator, and $c=B$ for the inner part of the denominator).
Step 4 · Evaluate each piece's limit
As $x\to\infty$: $\dfrac{1}{\sqrt x}\to0$, so the numerator tends to $\sqrt{1+0}=1$. For the denominator, $\dfrac Bx = \dfrac{\sqrt{x+\sqrt x}}{x}$; since $\sqrt{x+\sqrt x}$ grows like $\sqrt x$ while the denominator is $x$, this ratio behaves like $\dfrac{\sqrt x}{x}=\dfrac{1}{\sqrt x}\to0$. So the denominator tends to $\sqrt{1+0}+1 = 1+1=2$. Therefore:
$$ L = \frac{1}{2}. $$
Summary Table
| Quantity | Divided by $\sqrt x$ | Limit as $x \to \infty$ |
|---|---|---|
| $B=\sqrt{x+\sqrt x}$ | $\sqrt{1+\dfrac{1}{\sqrt x}}$ | $1$ |
| $A=\sqrt{x+B}$ | $\sqrt{1+\dfrac{B}{x}}$ | $1$ |
| $L=\dfrac{B}{A+\sqrt x}$ | — | $\dfrac12$ |
Additional Notes
- Because the final ratio only needed the leading-order limits of $B/\sqrt x\to1$ and $A/\sqrt x\to1$ (not their finer correction terms), the whole computation avoided any binomial or Taylor approximation — dividing by $\sqrt x$ was rigorous and exact at every step.
- A more refined analysis (tracking the next-order term) shows $A-\sqrt x \to \frac12$ not just in the limit but with $A \approx \sqrt x + \frac12 + O(x^{-1/2})$ for large $x$ — i.e., the nested radical $A$ eventually looks almost exactly like $\sqrt x + \frac12$, differing only by a term that vanishes as $x\to\infty$.
- The same rationalize-then-divide-by-$\sqrt x$ strategy handles any similarly nested radical limit, such as $\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}-\sqrt x$ with more layers of nesting — each additional layer only affects the size of the "inner" correction term, not the overall method.
Result
The value of the limit is: