lim⁡𝑥→0cos⁡(𝑥)−1𝑥x→0lim​ xcos(x)−1​ =A.- 1B.1C.Does not existD.2222​ ​ E.0

The limit in question is of the form 0/0, which is an indeterminate form. We can use L'Hopital's Rule, which states that the limit of a quotient of two functions as x approaches a certain value is equal to the limit of the quotients of their derivatives.

The derivative of cos(x) is -sin(x) and the derivative of x is 1. So, we have:

lim(x→0) [cos(x) - 1]/x = lim(x→0) [-sin(x)]/1

As x approaches 0, -sin(x) approaches 0. Therefore, the limit is 0.

So, the answer is E. 0.