The limit of sin(1/x) as x approaches 0 does not exist, that is, lim_{x→0} sin(1/x) is undefined. The limit formula of sin(1/x) when x→0 is given below:

lim_{x→0} sin(1/x) = undefined.

Let us learn how to find the limit of sin(1/x) when x approaches 0.

## Find the Limit of sin(1/x) when x→0

**Answer:** The function sin(1/x) has no defined limit when x→0.

*Proof:*

We will use the following theorem in order to prove the limit formula lim_{x→0} sin(1/x) = undefined.

**Theorem:** If lim_{x→c} f(x) = L, then for every sequence {x_{n}} converging to c the limit lim_{n→∞} f(x_{n}) always equals to L, that is, the limit will be unique.

Let f(x) = sin(1/x).

To show lim_{x→0} f(x) does not exist, consider two sequences {x_{n}} and {y_{n}}, both converging to 0, but f(x_{n}) and f(y_{n}) converge to different limits. Take

x_{n} = 1/nπ

y_{n} = $\dfrac{1}{(4n+1)\pi/2}$

Note that sin(nπ)=0 and sin (4n+1)π/2 = 1. Also, observe that x_{n}, y_{n} →0 when n tends to ∞. Now we calculate

lim_{n→∞} f(x_{n}) = lim_{n→∞} sin(nπ) = 0.

On the other hand,

lim_{n→∞} f(y_{n}) = lim_{n→∞} sin (4n+1)π/2 = 1.

Thus we deduce that both sequences f(x_{n}) and f(y_{n}) converge to different limits. Hence by the above theorem the limit of sin(1/x) when x→0 does not exist.

**ALSO READ:**

lim_{x→0} (e^{x}-1)/x = 1 | lim_{x→∞} sinx/x = 0 |

lim_{x→0} x/cosx = 0 | lim_{x→0} x/sinx = 1 |

lim_{x→0} tanx/x = 1 | lim_{x→∞} tanx/x = undefined |

Epsilon – delta definition of limit

## FAQs

### Q1: What is the limit of sin(1/x) as x approaches 0.

Answer: The limit of sin(1/x) as x approaches 0 is not defined.