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.
$\endgroup$3 Answers
$\begingroup$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:
$$\frac{\sin 5h}{3h} = \frac{\sin \theta}{3\frac{\theta}{5}}= \frac{5}{3}\frac{\sin \theta}{\theta}\stackrel{\theta to 0}{\longrightarrow}\frac{5}{3}$$
$\endgroup$ 1 $\begingroup$You probably know$$\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1\ .$$If you have sine of a multiple of $\theta$, for example as in your problem$$\lim_{\theta\to0}\frac{\sin5\theta}{3\theta}\ ,$$this makes it a bit harder. On the other hand if you have a multiple of $\theta$ in the denominator, it scarcely makes it harder at all because you can factor out the constant, e.g.,$$\lim_{\theta\to0}\frac{\sin\theta}{7\theta}=\frac17\lim_{\theta\to0}\frac{\sin\theta}{\theta}\ .$$So taking $\theta=5h$ would be the way to go IMHO.
$\endgroup$ $\begingroup$Let us first observe that $\lim\limits_{x \rightarrow 0} \frac {\sin x } {x} = 1$.
To show the above limiting value let us first try to give a bound for $\frac {\sin x} {x}$. Note that, $\sin x < x < \tan x$ for all $x \in (0,\frac {\pi} {2})$. Also for all $x \in (0,\frac {\pi} {2})$ we have $\sin x,\tan x > 0$. So we get from the above inequality that $$1 < \frac {x} {\sin x} <\sec x\ \text{for all}\ x \in \left (0, \frac {\pi} {2}\right).$$ or in other words
$$\cos x < \frac {\sin x} {x} <1\ \text{for all}\ x \in \left (0, \frac {\pi} {2}\right).$$
Now $\lim\limits_{x \rightarrow 0^+} \cos x =1$ and $\lim\limits_{x \rightarrow 0^+} 1 = 1$. So by Squeeze theorem we get $\lim\limits_{x \rightarrow 0^+} \frac{\sin x} {x} =1$.
To prove that $\lim\limits_{x \rightarrow 0^{-}} \frac {\sin x} {x} = 1$ use the inequality
$$\tan x < x < \sin x,\ \text {for all}\ x \in \left (-\frac{\pi} {2},0 \right ).$$
Now to evaluate $\lim\limits_{h \rightarrow 0} \frac {\sin 5h} {3h}$ first observe that for $h \neq 0$ $\frac {\sin 5h} {3h} = \frac {5} {3} \frac {\sin 5h} {5h}$. So
$$\lim\limits_{h \rightarrow 0} \frac {\sin 5h} {3h} = \frac {5} {3}\lim\limits_{h \rightarrow 0} \frac {\sin 5h} {5h} =\frac 5 3.$$ (Here you should observe that as $h \rightarrow 0,$ $5h \rightarrow 0$. So $\lim\limits_{h \rightarrow 0} \frac {\sin 5h} {5h} = \lim\limits_{5h \rightarrow 0} \frac {\sin 5h} {5h} = 1$).
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