# Limit of (cosx-1)/x as x approaches 0

The limit of (cosx-1)/x as x approaches 0 is equal to 0. Here, we will learn to find the limit of (cos(x)-1)/x when x→0. The limit formula of (cosx-1)/x as x tends to 0 is given by

limx→0 $\dfrac{\cos x -1}{x}$ = 0.

## Proof of Limit of (cosx-1)/x as x approaches 0

Prove that limx→0 $\dfrac{\cos x -1}{x}$ = 0.

Proof:

The following steps have to be followed in order to find the limit of cos(x)-1 divided by x as x approaches zero.

Step 1: Multiply both the numerator and the denominator of (cosx-1)/x by cosx+1. Thus, we obtain that

limx→0 $\dfrac{\cos x -1}{x}$ = limx→0 $\dfrac{(\cos x -1)(\cos x +1)}{x(\cos x +1)}$

= limx→0 $\dfrac{\cos^2 x -1}{x(\cos x +1)}$ by the formula (a-b)(a+b) = a2-b2.

Step 2: Applying the formula sin2x+cos2x=1, that is, cos2x -1 = -sin2x, the above limit will be equal to

= limx→0 $\dfrac{-\sin^2 x}{x(\cos x +1)}$

= – limx→0 $\dfrac{\sin x}{x}$ × limx→0 $\dfrac{\sin x}{\cos x +1}$ by the product rule of limits.

= -1 × $\dfrac{\sin 0}{\cos 0 +1}$as the limit of sinx/x is equal to 1 when x tends to 0.

= $-\dfrac{0}{1 +1}$

= 0.

So the value of the limit of (cosx-1)/x as x approaches 0 is equal to 0.

## Proof of limx→0 (cosx -1)/x = 0 by L’Hôpital’s rule

Let us now prove that the limit of (c0s(x)-1)/x as x approaches to 0 is equal to 0 by L’Hôpital’s rule. See that

(c0s0-1)/0 = (1-1)/0 = 0/0, so it is an intermediate form. Hence using the L’Hôpital’s rule, we get that

limx→0 $\dfrac{\cos x -1}{x}$

= limx→0 $\dfrac{\frac{d}{dx}(\cos x -1)}{\frac{d}{dx}(x)}$

= limx→0 $\dfrac{-\sin x}{1}$

= -sin0/1

= 0/1 as the value of sin0 is 0.

= 0.

Thus the limit of (cosx-1)/x when x approaches to 0 is equal to zero, and this is obtained by the L’Hôpital’s rule.

## FAQs

Q1: What is the limit of cosx-1/x as x approaches 0?

Answer: The limit of cosx-1/x as x approaches 0 is equal to zero.

Q2: What is the limit formula of (cosx-1)/x as x→0?

Answer: The limit formula of (cosx-1)/x as x→0 is equal to limx→0 (cosx -1)/x = 0.