# What is the limit of sinx / x as x goes to infinity?

Sep 28, 2015

By Squeeze Theorem, this limit is 0.

#### Explanation:

Since $\forall \epsilon > 0 , \implies 0 \le | \sin \frac{x}{x} | \le | \frac{1}{x} | \forall x \ge \frac{1}{\epsilon}$

and since $\lim 0 = 0 \mathmr{and} \lim \frac{1}{x} = 0$

By Squeeze Theorem $\implies \lim \sin \frac{x}{x} = 0$

Sep 28, 2015

${\lim}_{x \to \infty} \frac{\sin x}{x} = 0$

#### Explanation:

$0 \le \left\mid \frac{\sin x}{x} \right\mid \le \left\mid \frac{1}{x} \right\mid \to 0$ as $x \to \infty$ (also as $x \to - \infty$)

For any $\epsilon > 0$, we find $\left\mid \frac{\sin x}{x} \right\mid < \epsilon$ for all $x > \frac{1}{\epsilon}$.

$\frac{\sin x}{x}$ has some interesting properties and uses:

• ${\lim}_{x \to 0} \frac{\sin x}{x} = 1$

• $\frac{\sin x}{x} = 0 \setminus \iff \setminus x = k \pi \text{ for " k in ZZ " with } k \ne 0$

• $\frac{\sin x}{x}$ is an entire function. That is it is holomorphic at all finite points in the complex plane (taking its value at $x = 0$ to be $1$).

Hence by the Weierstrass factorisation theorem:

$\frac{\sin x}{x} = \left(1 - \frac{x}{\pi}\right) \left(1 + \frac{x}{\pi}\right) \left(1 - \frac{x}{2 \pi}\right) \left(1 + \frac{x}{2 \pi}\right) \left(1 - \frac{x}{3 \pi}\right) \left(1 + \frac{x}{3 \pi}\right) \ldots$

$\textcolor{w h i t e}{\frac{\sin x}{x}} = \left(1 - {x}^{2} / {\pi}^{2}\right) \left(1 - {x}^{2} / \left(4 {\pi}^{2}\right)\right) \left(1 - {x}^{2} / \left(9 {\pi}^{2}\right)\right) \ldots$

Hence the coefficient of ${x}^{2}$ in the Maclaurin expansion of $\frac{\sin x}{x}$ is:

$- {\sum}_{n = 1}^{\infty} \frac{1}{{n}^{2} {\pi}^{2}}$

But sin x = x/(1!)-x^3/(3!)+x^5/(5!)-..., so:

$\frac{\sin x}{x} = 1 - {x}^{2} / 6 + {x}^{4} / 120 - \ldots$

So: $\frac{1}{6} = {\sum}_{n = 1}^{\infty} \frac{1}{{n}^{2} {\pi}^{2}}$ and hence:

${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 2 = {\pi}^{2} / 6$

Finding this sum was the Basel Problem, set by Pietro Mengoli in 1644 and solved by Leonard Euler in 1734.

Aug 2, 2018

${\lim}_{x \to \infty} \sin \frac{x}{x} = 0$

#### Explanation:

What happens to our function when $x$ balloons up?

No matter what the input, $\sin x$ just oscillates between $0$ and $1$.

As the denominator gets larger and larger, we will be dividing by a larger number, which yields a smaller number.

Since the numerator stays relatively the same, and the denominator blows up, $\sin \frac{x}{x}$ will become infinitesimally small and approach zero.

Hope this helps!