Limit of x/sinx as x approaches 0

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

limx→0 (ex-1)/x = 1	limx→∞ sinx/x = 0
limx→0 (ax-1)/x = lna	limx→0 sinx/x
limx→0 sin(1/x)	limx→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

Sum Rule of Limits

Product Rule of Limits

Quotient Rule of Limits

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