trigonometry limits when x approaches zeroa) limx→0xsin⁡x=1b) limx→0sin⁡xx=1Is a) and b) true?Because if I try to apply this to limx→0sin⁡17xx, my answer is 17, not 1

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srasloavfv · Accepted Answer

Product of a number and an infinitesimal is still an infinitesimal. So 17×0→0. Actually an infinitesimal means a value approaching to zero. So if you multiply something (ofc not infinity) with an infinitesimal then you will get again an infinitesimal.Next notice that sin⁡{17x}x=17sin⁡{17x}17xThus limx→0sin⁡17xx=limx→017sin⁡{17x}17x=17

awalkbyfaithbzu6 · Accepted Answer

a) and b) are both truelimx→0sin⁡17xx=17limx→0sin⁡17x17x Lets operate the changee of variable y=17xy→0 as x→017limx→0sin⁡17x17x=17limy→0sin⁡yy=17×1=17

trigonometry limits when x approaches zero a) \(\displaystyle\lim_{{{x}\to{0}}}\frac{{x}}{{\sin{{x}}}}={1}\) b)

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