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Mathematics 17 Online
OpenStudy (maheshmeghwal9):

prove:

OpenStudy (maheshmeghwal9):

\[\lim_{x \rightarrow 0}\sin x/x\]

OpenStudy (anonymous):

L'Hopital rule.

OpenStudy (anonymous):

lim as x tends to zero for sinx/x = = lim as x tends to zero of cosx/1 L'Hopitals

OpenStudy (anonymous):

Technically that would do it for you, if you're allowed to assume L'hopital's.

OpenStudy (anonymous):

do you know that the derivative of sine is cosine? if so, then this is the same as asking what is \(\cos(0)\) if not, then you have to use the old "squeeze" theorem. any book will have this proof

OpenStudy (maheshmeghwal9):

k!

OpenStudy (anonymous):

absolutely no need for l'hopital here, \[\lim_{x\to 0}\frac{\sin(x)}{x}=\sin'(0)=\cos(0)=0\] by the definition of the derivative

OpenStudy (anonymous):

It is sandwich theorem for me :)), satellite73

OpenStudy (maheshmeghwal9):

k. Thanx!

OpenStudy (anonymous):

@satellite73 cos0=1

OpenStudy (maheshmeghwal9):

ya anhkhoava! u r ryt.

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