Need help finding the limit as x goes to infinity of function f(x) = (sin 10x)/(18x)

abel muroi

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Find the limit as x goes to positive infinity of this function

f(x) = (sin 10x)/(18x)


I know the answer is just 0, but i want to know the actual steps require to get the answer.

Can anyone show me the steps?
 
The way I'd tackle this problem is to use the squeeze theorem. If you need a refresher on the Squeeze Theorem, try this Khan Academy video. Think about what you know about the sine function. Specifically, what are the upper and lower bounds? Do those bounds vary as x changes? Based on this information, what two functions do you think would be appropriate to "sandwich" the main function between?
 
Find the limit as x goes to positive infinity of this function

f(x) = (sin 10x)/(18x)

I know the answer is just 0, but i want to know the actual steps require to get the answer.

Can anyone show me the steps?
This is where they expect you to remember that limit that they whizzed past near the beginning of the chapter on trig limits and derivatives; namely:

. . . . .limθ0\displaystyle \displaystyle \lim_{\theta\, \rightarrow\, 0}\, sin(θ)θ=0\displaystyle \dfrac{\sin(\theta)}{\theta}\, =\, 0

In this case, you'll need to make a clever adjustment to the "theta" portion. You've got 10x inside the sine, so you need to get 10x underneath, too. What happens if you multiply the fraction by 5/5 (as "useful" form of "1")?

. . . . .(sin(10x)18x)(55)=(sin(10x)10x)(59)\displaystyle \left(\dfrac{\sin(10x)}{18x}\right)\, \left(\dfrac{5}{5}\right)\, =\, \left(\dfrac{\sin(10x)}{10x}\right)\, \left(\dfrac{5}{9}\right)

Where does this lead? ;)
 
This is where they expect you to remember that limit that they whizzed past near the beginning of the chapter on trig limits and derivatives; namely:

. . . . .limθ0\displaystyle \displaystyle \lim_{\theta\, \rightarrow\, 0}\, sin(θ)θ=0\displaystyle \dfrac{\sin(\theta)}{\theta}\, =\, 0
;)

limit is as x+\displaystyle x\to +\infty
 
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