I just want lớn make sure I"m on the right path with the problem. The problem is as follows:

$$lim_x oinftyfracsin^2xx^2$$

I rewrote it as follows:

$$lim_x oinftyfrac(sin x)^2x^2$$

Now $sin(x)^2$ does oscillate as $x$ approaches infinity và therefore a limit does not exist. However it oscillates between the numbers $-1$ & $1$. Since the denominator would increase without bound và the numerator would only move between $-1$ and $1$, part of me wants khổng lồ say that the limit is zero.

Bạn đang xem: Sin2x = x

However a smarter part of me wants to lớn say that the limit does not exist due to the numerator. Could someone shed some light on this problem?

calculus algebra-precalculus limits trigonometry
nói qua
edited Jan 20, năm 2016 at 17:57

Martin Sleziak
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asked Jul 6, 2014 at 0:41

Irresponsible NewbIrresponsible Newb
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địa chỉ cửa hàng a comment |

2 Answers 2

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To solve the following question, recall the Squeeze Theorem:

So, as we know: $$0 le sin^2 x le 1$$

If we divide by $x^2$

$$0 le fracsin^2 xx^2 le frac1x^2$$

If we evaluate the limit from at either ends: $$lim_x o infty 0 = 0$$ $$lim_x o infty frac 1x^2 = 0$$

Therefore, by the squeeze theorem:

$$lim_x o infty fracsin^2 xx^2 = 0$$

giới thiệu
edited Dec 23, năm ngoái at 12:15
answered Jul 6, năm trước at 1:17

Varun IyerVarun Iyer
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What we have is as follows: for all $x>0$,$$0 leq fracsin^2 xx^2 leq frac 1x^2$$Now, chú ý that $lim_x o infty 0 = lim_x o infty frac 1x^2 = 0$. What theorem can we use here to lớn get the answer?

chia sẻ
answered Jul 6, năm trước at 0:44

Ben GrossmannBen Grossmann
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showroom a comment |

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