lim x --> 0 ((1-cos x)/ x^2)
How would I go about calculating this limit?
Thankyou
Starting from the 'fundamental limit'...
$\displaystyle \displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x}= 1$ (1)
... You first derive...
$\displaystyle \displaystyle \lim_{x \rightarrow 0} \frac{\sin^{2} x}{x^{2}} = \lim_{x \rightarrow 0} \frac{1-\cos^{2} x}{x^{2}} = \lim_{x \rightarrow 0} \frac{(1+\cos x)\ (1-\cos x)}{x^{2}}= 1$ (2)
... and from (2)...
$\displaystyle \displaystyle \lim_{x \rightarrow 0} (1+\cos x)\ \lim_{x \rightarrow 0} \frac{1-\cos x}{x^{2}} = 1 \implies \frac{1-\cos x}{x^{2}}= \frac{1}{2}$ (3)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$