The limit of cosx/x as x approaches infinity is equal to 0. The formula of lim_{x→∞} (cos x)/x is given below:

$\lim\limits_{x \to \infty} \dfrac{\cos x}{x}=0$.

Here we find the limit of (cos x)/x when x tends to infinity.

## Lim_{x→∞} (cos x)/x = 0 Proof

**Question: **What is the limit of cosx/x when x tends to infinity?

Answer: The limit of cosx/x is equal to 0 when x tends to infinity. |

**Explanation:**

The limit of cosx/x, when x→∞, will be computed using the Sandwich/Squeeze theorem of limits. We know that the value of cosx lies between -1 and 1, that is,

-1 ≤ cosx ≤ 1

For very large values of x, it follows that

$\dfrac{-1}{x} \leq \dfrac{\cos x}{x} \leq \dfrac{1}{x}$

Taking x→∞. we obtain that

lim_{x→∞} $\dfrac{-1}{x} \leq$ lim_{x→∞} $\dfrac{\cos x}{x} \leq$ lim_{x→∞} $\dfrac{1}{x}$

⇒ 0 ≤ lim_{x→∞} $\dfrac{\cos x}{x}$ ≤ 0

∴By the Sandwich theorem, the given limit

$\lim\limits_{x \to \infty} \dfrac{\cos x}{x} =0$

So the limit of cosx/x is equal to 0 when x tends to infinity, and this is proved by the Sandwich/Squeeze theorem of limits.

**Read Also:** Limit of sinx/x when x→∞

More Limits

lim_{x→0} sin(√x)/x | lim_{x→0} sin(x^{2})/x |

lim_{x→0} sin3x/sin2x | lim_{x→}0 x/sinx |

lim_{x→0} tanx/x | lim_{x→0}(cosx-1)/x |

## FAQs

### Q1: What is the limit of cosx/x when x tends to infinity?

Answer: The limit of cosx/x^{ }is equal to 0 when x tends to ∞, that is, lim_{x→∞} cosx/x =0.