Question

lim of [sin(pie+∆x)]/∆x as ∆x→0

Answer 1

so basically you are trying to find the derivative of the function sin(x)

Answer 2

?

Answer 3

that isn't even the definition of the derivative

Answer 4

Actually it is.

Recall first that the derivative of f at a is defined as the limit as h -> 0 of
(f(x+h) - f(x))/h

Here f = sin, and x = pi

The ratio is

(sin(pi+h) - sin(pi))/h = sin(pi+h)/h, as sin(pi) = 0.

So the limit of this ratio as h -> 0 is d(sin x)/dx evaluated at x = 0.

Now, the derivative of sin(x) is cos (x). Hence your limit is equal to ...

Answer 5

he's missing -f(c)

Answer 6

\[\lim _{\Delta x \rightarrow0} \sin(\pi+\Delta x) / \Delta x\]

Answer 7

if he's doing the definition it should be\[\frac{f(c+\Delta x)-f(c)}{\Delta x}\]

Answer 8

@willking: yes
@outkast: no, because sin(pi) = 0.

Answer 9

In your notation f(c) = sin(pi) = 0.

Answer 10

ohh but pshh the question should actually state that... it shouldn't just totally detract the statement from the definition

Answer 11

So ... @willking9 ... go back to my main answer. Do you follow and can you now see the answer?

Answer 12

also you cannot assume that d/dx[sin]=cos

Answer 13

If i were asked to use the definition and just wrote sin(x)=cos(x) .... i'd get it wrong

Answer 14

You need a trig identity

Answer 15

How do you know you can't assume the derivative of sin?

But let's assume you're right. Show us how smart you are by showing us the trig identity in action:

Simplify and evaluate the difference quotient

(sin(x+h) - sin(x))/h

as h --> 0

Answer 16

Because the point of using the definition of the derivative is to find out the derivative Without an assumption

Answer 17

Ok ... so, show us how it's done, enough postering. What have you got?

Answer 18

I'll do it then.

sin a - sin b = 2 sin( (a-b)/2 ) . cos( (a+b)/2 )

So ( sin(pi + h) - sin(pi) ) / h = 2 sin(h/2) . cos (pi + h) / h
= cos(pi+h) . [ sin(h/2)/(h/2) ]

Now, it is case that the limit as x -> 0 of sin(x)/x = 1, so this last term in square brackets, as h -> 0, is 1, while the limit of cos(pi+h) is cos(pi)

Hence the limit as h -> 0 of
( sin(pi + h) - sin(pi) ) / h = cos(pi).1 = -1

**************

Answer 19

or like i was saying do the proof of d/dx[sin(x)]=cos(x) and evaluating at c
in which you would use
the Trig Identity

sin(c+h)=sin(c)cos(h)+cos(c)sin(h)

Answer 20

Just use the fact that \[\sin(\pi+\Delta x)=-\sin(\Delta x)\]

then \[\lim _{\Delta x \rightarrow0} \frac{\sin(\pi+\Delta x)}{\Delta x}=\lim _{\Delta x \rightarrow0} \frac{-\sin(\Delta x)}{\Delta x}=-1\]

Answer 21

Ah, that's elegant.

Answer 22

i was told to use the trig identity that states that sin(A+B)=sinAcosB+cosAsinB

Answer 23

\[\sin(\pi+\Delta x)=\sin(\pi)\cos(\Delta x)+\cos(\pi)\sin(\Delta x)\]

\[=0\cos(\Delta x)+(-1)\sin(\Delta x)=-\sin(\Delta x)\]