Limit of cos(x/sqrt(n)) raised to the n
For a fixed real number $x$, evaluate the limit: $$ L = \lim_{n \to +\infty} \cos^n\left(\frac{x}{\sqrt{n}}\right). $$
Explanation of Method
For fixed $x$, as $n\to\infty$ the argument $\dfrac{x}{\sqrt n}\to 0$, so $\cos\left(\dfrac{x}{\sqrt n}\right)\to\cos(0)=1$ while the exponent $n\to\infty$, giving the indeterminate form $1^\infty$. The strategy is to expand $\cos$ with its Taylor series around $0$ to express the base as $1+\dfrac{a}{n}+o\!\left(\dfrac1n\right)$ for a specific constant $a$, and then use the defining property $\left(1+\dfrac{a}{n}\right)^n \to e^{a}$ as $n\to\infty$.
- Expand $\cos(z)$ with its Taylor series near $z=0$, where $z=\dfrac{x}{\sqrt n}$.
- Substitute to express the base as $1 - \dfrac{x^2}{2n} + O\!\left(\dfrac{1}{n^2}\right)$.
- Apply the standard limit $\left(1+\dfrac{a}{n}\right)^n \to e^{a}$.
Step 1 · Expand cosine near $0$
The Taylor series for $\cos(z)$ around $z=0$ is:
$$ \cos(z) = 1 - \frac{z^2}{2} + O(z^4). $$
As $n\to\infty$, $z = \dfrac{x}{\sqrt n} \to 0$ for any fixed $x$, so this expansion applies.
Step 2 · Substitute $z = \frac{x}{\sqrt n}$
Since $z^2 = \dfrac{x^2}{n}$:
$$ \cos\left(\frac{x}{\sqrt n}\right) = 1 - \frac{x^2}{2n} + O\!\left(\frac{1}{n^2}\right). $$
Step 3 · Raise to the power $n$
We now need:
$$ \cos^n\left(\frac{x}{\sqrt n}\right) = \left(1 - \frac{x^2}{2n} + O\!\left(\frac{1}{n^2}\right)\right)^n. $$
Recall the defining property of the exponential function:
$$ \left(1 + \frac{a}{n}\right)^n \longrightarrow e^{a} \quad \text{as } n \to \infty, \quad \text{for any fixed constant } a. $$
Here $a = -\dfrac{x^2}{2}$ (the $O(1/n^2)$ correction term, once multiplied out through the $n$-th power, contributes a factor that tends to $1$ and does not affect the limit). Therefore:
$$ \cos^n\left(\frac{x}{\sqrt n}\right) \longrightarrow e^{-\frac{x^2}{2}} \quad \text{as } n \to \infty. $$
Summary Table
| Quantity | Expansion / value | Behavior as $n \to \infty$ |
|---|---|---|
| $\dfrac{x}{\sqrt n}$ | — | $\to 0$ |
| $\cos\left(\dfrac{x}{\sqrt n}\right)$ | $1-\dfrac{x^2}{2n}+O\!\left(\dfrac1{n^2}\right)$ | $\to 1$ |
| $\cos^n\left(\dfrac{x}{\sqrt n}\right)$ | — | $\to e^{-x^2/2}$ |
Additional Notes
- This limit is the essential computation behind why the binomial distribution converges to the Gaussian (normal) distribution — the function $e^{-x^2/2}$ is, up to a normalizing constant, the density of the standard normal distribution.
- The key structural fact used is that $\left(1+\frac{a}{n}\right)^n \to e^a$ for any fixed real $a$; any base that can be written as $1+\frac{a}{n}+o\!\left(\frac1n\right)$ raised to the $n$-th power converges to $e^a$ by the same reasoning.
- The variable $x$ is held fixed throughout — the result $e^{-x^2/2}$ is a function of $x$, describing how the limiting expression depends on this fixed parameter as $n\to\infty$.
Result
The value of the limit is: