Suppose that the task is to compute the limits of these functions as t→0.
- limt→0tsint. It is a well-known limit. It is possible to compute this limit by using geometric considerations or by using L'Hôpital's rule. Namely, (sint)′=cost and t′=1, where ′ denotes the derivative. Therefore, limt→0tsint=limt→01cost=1
- limt→0t1−cost. Using trigonometric formulae, one receives: 1−cost=2sin22t. We use the limit from the previous exercise and get: limt→0t1−cost=limt→024t2sin22t4t=0.
- limt→0tet−1. It is also a well-known limit. It is possible to compute the limit using different considerations. We use L'Hôpital's rule: (et−1)′=et,t′=1. Therefore: limt→0tet−1.=limt→0et=1.
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