Suppose that the task is to compute the limits of these functions as "t\\rightarrow0".
- "\\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, "(sin\\,t)'=cos\\,t" and "t'=1", where "'" denotes the derivative. Therefore, "\\lim_{t\\rightarrow 0}\\frac{sin\\, t}{t}=\\lim_{t\\rightarrow 0}\\frac{cos\\, t}{1}=1"
- "\\lim_{t\\rightarrow 0}\\frac{1-cos\\, t}{t}". Using trigonometric formulae, one receives: "1-cos\\,t=2\\,sin^2\\,\\frac{t}{2}". We use the limit from the previous exercise and get: "\\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."
- "\\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: "(e^t-1)'=e^t,\\quad t'=1". Therefore: "\\lim_{t\\rightarrow 0}\\frac{e^t-1}{t}.=\\lim_{t\\rightarrow 0}e^t=1."
Comments
Leave a comment