Question

what is the Limit as x approaches negative infinity sinx/x...

Answer 1

\[\lim_{x\to-\infty}\frac{\sin x}{x}\]
\(\sin x\) is bounded between -1 and 1, so as x increases arbitrarily, \(\sin x\) will oscillate between -1 and 1.

So for our purposes here, you can say \(\sin x=\pm1\) for large \(x\):
\[\lim_{x\to-\infty}\frac{\pm1}{x}\]

Answer 2

Sorry, *decreases* arbitrarily.

Answer 3

ok i got that sinx/x equals 1 so I assumed that the answer was one but my book says the answer is 0

Answer 4

Ah, you're confusing it with the other limit, \(\displaystyle\lim_{x\to0}\dfrac{\sin x}{x}=1\). For this problem, we're given a limit at infinity.

Answer 5

yea i started to piece that together too thats why i asked the question but I don't understand how the book got 0..... I dont understand how to get a value period

Answer 6

Using my reasoning, that
\[\lim_{x\to-\infty}\frac{\sin x}{x}=\lim_{x\to-\infty}\frac{\pm1}{x}\]
as \(x\) gets bigger and bigger, the value of \(\dfrac{1}{x}\) approaches 0. So the limit is 0.

Answer 7

ok so what if 1-cosx/x^2 was also reaching infinity would it also be zero

Answer 8

Yes. \(\cos x\) behaves similarly to \(\sin x\), in the sense that it's also bounded between -1 and 1. So the numerator is a constant, which makes the denominator the determining factor of the limit's value.

Answer 9

ok...because my teacher plugs in like 100,000,00 into the denominator to prove the point that the problem is get infinitely closer to 0 so it makes the the answer zero

Answer 10

It's the same reasoning, really.

Answer 11

oh ok.... yea cuz i was confused when I was suppose to use that idea but now i see thats

Answer 12

that only occurs with infinity

Answer 13

Yeah, usually you can apply this sort of reasoning to determine limits at infinity.

Answer 14

ok....thanks

Answer 15

You're welcome