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.

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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
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edited Jan 20, năm 2016 at 17:57

Martin Sleziak
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\$\$

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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?

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answered Jul 6, năm trước at 0:44

Ben GrossmannBen Grossmann
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