# What is the limit lim_(x->0)(cos(x)-1)/x?

Oct 11, 2014

${\lim}_{x \to 0} \frac{\cos \left(x\right) - 1}{x} = 0$. We determine this by utilising L'hospital's Rule.

To paraphrase, L'Hospital's rule states that when given a limit of the form lim_(x→a)f(x)/g(x), where $f \left(a\right)$ and $g \left(a\right)$ are values that cause the limit to be indeterminate (most often, if both are 0, or some form of ∞), then as long as both functions are continuous and differentiable at and in the vicinity of $a ,$ one may state that

lim_(x→a)f(x)/g(x)=lim_(x→a)(f'(x))/(g'(x))

Or in words, the limit of the quotient of two functions is equal to the limit of the quotient of their derivatives.

In the example provided, we have $f \left(x\right) = \cos \left(x\right) - 1$ and $g \left(x\right) = x$. These functions are continuous and differentiable near $x = 0 , \cos \left(0\right) - 1 = 0 \mathmr{and} \left(0\right) = 0$. Thus, our initial f(a)/g(a)=0/0=?.

Therefore, we should make use of L'Hospital's Rule. $\frac{d}{\mathrm{dx}} \left(\cos \left(x\right) - 1\right) = - \sin \left(x\right) , \frac{d}{\mathrm{dx}} x = 1$. Thus...

${\lim}_{x \to 0} \frac{\cos \left(x\right) - 1}{x} = {\lim}_{x \to 0} \frac{- \sin \left(x\right)}{1} = - \sin \frac{0}{1} = - \frac{0}{1} = 0$