Limit of (1-cosx)/x^2 as x approaches 0

The limit of (1-cosx)/x^2 as x approaches 0 is equal to 1/2. That is, lim_x→0 (1-cosx)/x² =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)/x² =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=2sin²x.

= 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 $\dfrac{\sin^2 h}{h^2}$

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

= $\dfrac{1}{2}$ × 1² 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 $\dfrac{\cos x -1}{x}$

lim_x→0 $\dfrac{\sin x}{x}$

lim_x→0 $\dfrac{\cos x}{x}$

lim_x→0 $\dfrac{\tan x}{x}$

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