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.
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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
edited Jan 20, năm 2016 at 17:57
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asked Jul 6, 2014 at 0:41
Irresponsible NewbIrresponsible Newb
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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:
$$lim_x o infty fracsin^2 xx^2 = 0$$
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?
answered Jul 6, năm trước at 0:44
Ben GrossmannBen Grossmann
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