The value of the limit of x/sinx is equal to 1 when x tends to 0. In this post, we will learn how to find the limit of $\frac{x}{\sin x}$ when x approaches 0.

## lim x/sinx when x approaches 0 Formula

The formula of the limit of x/sinx when x tends to zero is given below:

$\lim\limits_{x \to 0} \dfrac{x}{\sin x}=1$.

## lim_{x→0} x/sinx = 1 Proof

We will use the well-known limit formula for sine functions which is $\lim\limits_{x \to 0} \dfrac{\sin x}{x}=1$ **…(I)**

Note that

$\lim\limits_{x \to 0} \dfrac{x}{\sin x}$ $=\lim\limits_{x \to 0} \dfrac{1}{\frac{\sin x}{x}}$

= $\dfrac{1}{\lim\limits_{x \to 0} \frac{\sin x}{x}}$

= $\dfrac{1}{1}$ by the above limit formula **(I)**

= 1.

So the limit of x/sinx when x tends to 0 is equal to 1.

**ALSO READ:**

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

lim_{x→0} (a^{x}-1)/x = lna | lim_{x→0} sinx/x |

lim_{x→0} sin(1/x) | lim_{x→0} cos(1/x) |

**Alternative Proof:**

lim_{x→0} x/sinx = 1 can be proved using the L’Hôpital’s rule.

Note that lim_{x→0} x/sinx = 0/sin0 = 0/0, so it is an indeterminate form and we can use L’Hôpital’s rule to find its limit. Now,

$\lim\limits_{x \to 0} \dfrac{x}{\sin x}$

= $\lim\limits_{x \to 0} \dfrac{\frac{d}{dx}(x)}{\frac{d}{dx}(\sin x)}$

= $\lim\limits_{x \to 0} \dfrac{1}{\cos x}$

= $\dfrac{1}{\cos 0}$

= $\dfrac{1}{1}$ as the value of cos0 is 1.

= 1.

So the limit of x/sinx is equal to 1 when x approaches zero, and this is proved by the L’Hôpital’s rule.

**More Reading:** Solved Problems on Exponential Limits

## FAQs

**Q1: What is the limit of x/sinx when x approaches 0?**

**Answer:** The limit of x/sinx is equal to 1 when x approaches to 0.