The limit of (1-cosx)/x^2 as x approaches 0 is equal to 1/2. That is, lim_{x→0} (1-cosx)/x^{2} =1/2. This limit can be computed using the formula

lim_{x→0} $\dfrac{\sin x}{x}$ = 1 …(∗)

## Prove that lim_{x→0} (1-cosx)/x^{2} =1/2

We have

lim_{x→0} $\dfrac{1-\cos x}{x^2}$

= lim_{x→0} $\dfrac{2 \sin^2 \frac{x}{2}}{x^2}$ using the formula 1-cos2x=2sin^{2}x.

= lim_{x→0} $\dfrac{2 \sin^2 \frac{x}{2}}{(\frac{x}{2})^2 \times 4}$

= $\dfrac{1}{2}$ lim_{x→0} $\dfrac{\sin^2 \frac{x}{2}}{(\frac{x}{2})^2}$

[Let x/2 =h, so h→0 as x→0]

= $\dfrac{1}{2}$ lim_{h→0} $\dfrac{\sin^2 h}{h^2}$

= $\dfrac{1}{2} \big( \lim\limits_{h \to 0} \dfrac{\sin h}{h} \big)^2$

= $\dfrac{1}{2}$ × 1^{2} by the above formula (∗)

= 1/2.

So the limit of (1-cosx)/x^2 is equal to 1/2 when x tends to 0.

**More Limits:**

lim_{x→0} $\dfrac{\cos x -1}{x}$

## FAQs

### Q1: What is the limit of (1-cosx)/x^2 when x tends to 0?

Answer: The limit of (1-cosx)/x^{2} when x tends to 0 is equal to 1/2, that is, lim_{x→0} (1-cosx)/x^{2} =1/2.