# calculus help again

if f'(x)=cos x and g'(x)=1 for all x, and if f(0)=g(0)=0, then the
limit x--->0 fo function f(x)/g(x)=

okay, so f(x)=sinx
g(x)=x
and the f(0)=g(0)=0 is also satisfied and equals o.

so the limit x-->o of sinx/x=

is the answer nonexistent according to the conditions of this problem because,

in my math book, it says the if you take the limit x-->o of sin(x)/x=1

are they like two different things. this is a mutliple choice question so o and nonexistent are included in the choices.

i hope you guys can understand what i'm saying. do i go with the book and say the answer is one or should i say the limit does not exist

You need to use l'hopitals rule to understand that its an indefinite derivative or something complicated like that.

You just need to know that sin (h) / h = 1 and cos (h) - 1 / h = 0

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asked by david

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