Question

Need help with calculus problem.
FAN & MEDAL
problem in comments

Answer 1

Compute
\[\lim_{x \rightarrow \infty} \frac{ \sin x }{ x }\]

Answer 2

try to use sandwich rule, which is basically rule state that if your function is between two other functions with the same limit, then your function will have this limit

we can see that range of sinx is -1 and 1, so we can see \(-\dfrac{1}{x}\leq\dfrac{\sin x}{x}\leq\dfrac{1}{x}\)

Now, what's \(\large \displaystyle \lim_{x\rightarrow \infty}-\dfrac{1}{x}\) and \(\large \displaystyle \lim_{x\rightarrow \infty}\dfrac{1}{x}\)?

Answer 3

0?

Answer 4

yep, and according to sandwich rule, \(\displaystyle \large\lim_{x\rightarrow\infty}\dfrac{\sin x}{x} = ~?\)

Answer 5

0

Answer 6

that's correct. does that make sense??

Answer 7

and i think it's called the squeeze principal
and yes it does

Answer 8

yep, basically another name for it. sandwich rule sound more fun to use lol

Answer 9

welcome, by the way

Answer 10

http://larswiki.larseighner.com/uploads/Main/pinchsinthetaovertheta.svg

The area of the triangle OMT is greater than area of the sector theta is greater than the area of the triangle OMP. The purple line is tan theta. That is the setup. (All of this is "near" theta = 0.)

PS: the "squeezing" used to be call "the pinching theorem."

Answer 11

what are you trying to say? @LarsEighner

Answer 12

Since OM = 1 (this is the unit circle, the area of the big triangle is (1/2) tan theta.
The area of the sector theta is \( {1 \over 2} r^2 \theta= {1 \over 2} \theta \) since this is the unit circle and r=1.

The area of OMP is \({1\over 2} \sin \theta \cos \theta \)

so
\( \huge {1 \over 2} \sin \theta \cos \theta \le {1 \over 2} \theta \le {1 \over 2}\tan \theta \)

Answer 13

Multiply by 2 and divide by \(\sin \theta \)

\(\huge \cos \theta \le \frac{\theta}{\sin\theta} \le \frac{1}{\cos\theta}\)

\(\huge \cos \theta \ge \frac{\sin \theta}{\theta} \ge \frac{1}{\cos\theta}\)

Now actually the inequalities are reversed for small neqative theta where sin is negative, but as will be seen that case is only trivially different.

\( \huge \displaystyle {\lim_{\theta \rightarrow 0} \cos \theta = 1} \)

\( \huge \displaystyle \lim_{\theta \rightarrow 0} {1 \over {\cos \theta}} = 1\)

so \( {{\sin \theta|} \over \theta} \) is pinced/trapped/squeezed.

Answer 14

As can be seen, the sense of the inequalities does not really matter so long as \({{\sin\theta}\over \theta} \) is trapped between cos theta and its reciprocal. So for small negative theta where sin theta is negative the inequalities are reversed, but still trap \({{\sin\theta}\over \theta} \)