Question

Evalute limits involving the by constructjng their respective table of values

1.lim t -> 0 t/(sin t)

2.lim t -> 0 (sin t)/(2t)•(1 - cot t)/t

3.lim t -> 0 (e ^ t - 1/t)

Answer 1

1. $\lim_{t\rightarrow0}\frac{t}{sin\,t}=\lim_{t\rightarrow0}(\frac{sin\,t}{t})^{-1}=1.$ It contains the well-known limit $\frac{sin\,t}{t}$ The values of function $\frac{t}{sin\,t}$ near are: $t=0.05,\frac{t}{sin\,t}\approx1.0004;$ $t=0.03,\frac{t}{sin\,t}\approx1.0002;$ $t=0.01;\frac{t}{sin\,t}\approx1.00002.$ From the latter we can conclude that the limit is 1. Another way to compute the limit is to use L'Hôpital's rule.
2. $\lim_{t\rightarrow0}\frac{sin\,t}{2t}\frac{1- cot\,t}{t}=+\infty$ . The first fraction is the well-known limit multiplied by $\frac{1}{2}$. The second fraction can be rewritten in the form: $\frac{1-cot\,t}{t}=\frac{1-\frac{cos\,t}{sin\,t}}{t}=\frac{cos\,t-sin\,t}{t\,sint}$. This fraction approaches infinity (the term $cos\,t-sin\,t$ approaches 1 and the term $t\,sint$ approaches 0). Another way to compute the limit is to use L'Hôpital's rule.
3. $\lim_{t\rightarrow0}\frac{e^t-1}{t}=1$. It is the well-known known limit $\frac{e^t-1}{t}$ . The values of function $\frac{e^t-1}{t}$ near are: $t=0.05$, $\frac{e^t-1}{t}\approx1.0152;$ $t=0.03,\frac{e^t-1}{t}\approx1.0152;$ $t=0.01,$ $\frac{e^t-1}{t}\approx1.0050.$ Another way to compute the limit is to use L'Hôpital's rule.