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

Answer 1

#lim_(x->0) (cos(x)-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(a)# and #g(a)# 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))#

Stated differently, the derivative of two functions has a limit equal to the limit of the quotient of its derivatives.

In the example provided, we have #f(x)=cos(x)-1# and #g(x)=x#. These functions are continuous and differentiable near #x=0, cos(0) -1 =0 and (0)=0#. Thus, our initial #f(a)/g(a)=0/0=?.#

Therefore, we should make use of L'Hospital's Rule. #d/dx (cos(x) -1)=-sin(x), d/dx x=1#. Thus...

#lim_(x->0) (cos(x)-1)/x = lim_(x->0)(-sin(x))/1 = -sin(0)/1 = -0/1 = 0#

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Answer 2

Ezra Adams

The limit of (cos(x) - 1) / x as x approaches 0 is 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

What is the limit #lim_(x-&gt;0)(cos(x)-1)/x#?

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