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OpenStudy (anonymous):
L'Hopital's rule will work here.
OpenStudy (anonymous):
Or, more simply, use the rule that the limit of the product is the product of the limits.\[\lim_{x→0 }\frac{ (\sin⁴ x)}{x}=\lim_{x→0 }\sin^3x \lim_{x→0 }\frac{ (\sin x)}{x}=0\]
OpenStudy (anonymous):
Recall that \[\lim_{x \rightarrow 0}\frac{sinx}{x}=1\]
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
so should it be 1⁴?
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OpenStudy (anonymous):
No. How did you get that?
OpenStudy (anonymous):
well the identity gives us one
OpenStudy (anonymous):
Yeah, but the other limit \[\lim_{x \rightarrow0}\sin^3x =0\] is also a factor, and sends the limit of the product to zero. One times zero is zero.
OpenStudy (anonymous):
oh I follow now
OpenStudy (anonymous):
No sweat. Do math every day.
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OpenStudy (anonymous):
thank you, I just didn't know I could separate to make that identity thanks a lot for your help
OpenStudy (anonymous):
Just remember the rules, and figure out how to make the problem look like something that you can solve.