Question #320180

1.(sint)/t


2.(1 - cos t)/t


3.(e ^ t - 1)/t

1
Expert's answer
2022-03-31T17:07:48-0400

Suppose that the task is to compute the limits of these functions as t0t\rightarrow0.

  1. limt0sintt.\lim_{t\rightarrow 0}\frac{sin\, t}{t}. 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(sin\,t)'=cos\,t and t=1t'=1, where ' denotes the derivative. Therefore, limt0sintt=limt0cost1=1\lim_{t\rightarrow 0}\frac{sin\, t}{t}=\lim_{t\rightarrow 0}\frac{cos\, t}{1}=1
  2. limt01costt\lim_{t\rightarrow 0}\frac{1-cos\, t}{t}. Using trigonometric formulae, one receives: 1cost=2sin2t21-cos\,t=2\,sin^2\,\frac{t}{2}. We use the limit from the previous exercise and get: limt01costt=limt02sin2t2t24t4=0.\lim_{t\rightarrow 0}\frac{1-cos\, t}{t}=\lim_{t\rightarrow 0}2\frac{sin^2\, \frac{t}{2}}{\frac{t^2}{4}}\frac{t}4=0.
  3. limt0et1t.\lim_{t\rightarrow 0}\frac{e^t-1}{t}. It is also a well-known limit. It is possible to compute the limit using different considerations. We use L'Hôpital's rule: (et1)=et,t=1(e^t-1)'=e^t,\quad t'=1. Therefore: limt0et1t.=limt0et=1.\lim_{t\rightarrow 0}\frac{e^t-1}{t}.=\lim_{t\rightarrow 0}e^t=1.

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