I found a Question from the textbook which is not well explained…or not at all explained. And it would be interesting to know how to solve it. Or perhaps I am simply that bad at understanding squeeze theorem.
- If you want to evaluate $\lim _ { h \rightarrow 0 } \frac { \sin 5 h } { 3 h } ,$ it is a good idea to rewrite the
limit in terms of the variable (choose one):
$\begin{array} { l l l } { \text { (a) } \theta = 5 h } & { \text { (b) } \theta = 3 h } & { \text { (c) } \theta = \frac { 5 h } { 3 } } \end{array}$
Basically: for the limit as h approaches $0$, in the equation: $\sin(5h/3h)$, how should you rewrite the limit in terms of the variable theta?
For clarity's sake, I don't care much for the answer itself. It's a "completion" grade anyway. I am, however, interested in what logic goes into determining how you rewrite the limit.
Best Answer
The logic behind is that $\lim_{x\to 0}\frac{\sin x}{x}= 1$ is known.
So, one would like to transform the expression to contain this limit which excludes the "problem with the sine".
Hence, $\theta = 5h$ would be a good choice.
Now, you will see how this transforms the limit in a nice way:
$$\frac{\sin 5h}{3h} = \frac{\sin \theta}{3\frac{\theta}{5}}= \frac{5}{3}\frac{\sin \theta}{\theta}\stackrel{\theta to 0}{\longrightarrow}\frac{5}{3}$$