anonymous
  • anonymous
Why in this case does lim t -> 0 [8t/sin(8t)] =1 ? lim t ->0 [sin(3t)/3t] * lim t -> 0 [8t/sin(8t)] BUT in this case lim x -> 0 x/sin(7x) we have to invert it?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
The function you're asking about is the reciprocal of \[\frac{\sin8t}{8t}\] What's the limit of this as \(t\to0\)?
anonymous
  • anonymous
So even as the reciprocal the limit = 1?
anonymous
  • anonymous
At least in this case, yes. The reasoning involves use of the squeeze theorem, I think, but another way is due to the behavior of the functions near 0. If you were to plot \(\sin 8x\) and \(8x\), you'd see that \(\sin8x\approx8x\) for \(x\) near 0. So you have \[\lim_{x\to0}\frac{\sin8x}{8x}=\lim_{x\to0}\frac{8x}{\sin8x}=\lim_{x\to0}\frac{8x}{8x}=1\]

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anonymous
  • anonymous
I see that the graphs are very close near 0; what's a case where this wouldn't be true?
anonymous
  • anonymous
I guess I don't know how I would have known that.
anonymous
  • anonymous
why do you say "lim x -> 0 x/sin(7x) we have to invert it?"
anonymous
  • anonymous
you can multiply by 7/7 and get your answer 1/7 since lim 7x/sin7x = 1
anonymous
  • anonymous
None of the resources/profs said that it was true either way. I thought the sin function had to be in the numerator.
anonymous
  • anonymous
i think about the taylor series of sinx about x=0 sinx = x - x^3/3! +.. when x really goes to 0 the approx sinx=x works very good and so for sin7x = 7x so it doesnt really matter if it is up or down

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