Limit of the x-th root of cos(sqrt(x))
Evaluate the limit: $$ L = \lim_{x \to 0^+} \sqrt[x]{\cos\left(\sqrt{x}\right)}. $$
Explanation of Method
Since $\sqrt{x}$ requires $x \ge 0$, this limit is understood as $x\to 0^+$. As $x\to0^+$, $\cos(\sqrt x)\to\cos(0)=1$ while the root index $\frac1x\to+\infty$, giving the indeterminate form $1^\infty$. Rewriting the $x$-th root as $e^{\ln(f(x))/x}$ turns the problem into evaluating an exponent limit using a Taylor expansion, then exponentiating at the end.
- Rewrite $\sqrt[x]{f(x)}$ as $e^{\ln(f(x))/x}$.
- Expand $\cos(\sqrt x)$ with its Taylor series to order $x$.
- Take the logarithm and simplify using $\ln(1+u)\approx u$ for small $u$.
- Evaluate the resulting exponent limit and exponentiate.
Step 1 · Rewrite the root as an exponential
For $f(x) > 0$ near the limit point, the $x$-th root can be written as:
$$ \sqrt[x]{f(x)} = f(x)^{1/x} = e^{\frac{\ln(f(x))}{x}}. $$
With $f(x) = \cos(\sqrt x)$:
$$ L = \lim_{x \to 0^+} e^{\frac{\ln\left(\cos(\sqrt x)\right)}{x}}. $$
Step 2 · Expand $\cos(\sqrt x)$ near $x=0$
Let $z = \sqrt x$, so $z^2 = x$. Using the Taylor series $\cos(z) = 1 - \dfrac{z^2}{2} + O(z^4)$:
$$ \cos\left(\sqrt x\right) = 1 - \frac{x}{2} + O(x^2). $$
Step 3 · Take the logarithm
Since $\cos(\sqrt x) \to 1$ as $x\to 0^+$, use the small-argument approximation $\ln(1+u)\approx u$ with $u = -\dfrac{x}{2}+O(x^2) \to 0$:
$$ \ln\left(\cos(\sqrt x)\right) = \ln\left(1 - \frac{x}{2} + O(x^2)\right) \approx -\frac{x}{2} + O(x^2). $$
Step 4 · Evaluate the exponent and exponentiate
Dividing by $x$:
$$ \frac{\ln\left(\cos(\sqrt x)\right)}{x} \approx \frac{-\frac{x}{2}+O(x^2)}{x} = -\frac{1}{2} + O(x) \longrightarrow -\frac{1}{2} \quad \text{as } x \to 0^+. $$
Substituting back into the exponential form:
$$ L = \lim_{x\to0^+} e^{\frac{\ln(\cos(\sqrt x))}{x}} = e^{-\frac12}. $$
This can also be written as:
$$ L = e^{-1/2} = \frac{1}{\sqrt{e}}. $$
Summary Table
| Quantity | Series / approximation | Limit as $x \to 0^+$ |
|---|---|---|
| $\cos(\sqrt x)$ | $1-\dfrac{x}{2}+O(x^2)$ | $1$ |
| $\dfrac{\ln(\cos\sqrt x)}{x}$ | $-\dfrac12+O(x)$ | $-\dfrac12$ |
| $L=\sqrt[x]{\cos\sqrt x}$ | $e^{-1/2}$ | $\dfrac{1}{\sqrt e}$ |
Additional Notes
- Because $\sqrt{x}$ is only real for $x\ge0$, the one-sided limit $x\to0^+$ is the only version of this problem that makes sense — there is no two-sided limit to consider.
- The log-and-exponentiate technique used here is the standard method for any $1^\infty$, $0^0$, or $\infty^0$ indeterminate form, since it converts a power into a product that is often much easier to expand with a Taylor series.
- Only the linear (order-$x$) term of the expansion of $\cos(\sqrt x)$ mattered, since it is divided by $x$ before taking the limit — any $O(x^2)$ correction vanishes in that final step.
Result
The value of the limit is: