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Find the limit?

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2 Answers

Use the trig identity sin(x+h) = sin(x)cos(h) + cos(x)sin(h), so that
(sin(x+h)-sin(x))/h = (sin(x)(cos(h) - 1) + cos(x)sin(h))/h.
Now take the limit
limh→0 (sin(x)(cos(h) - 1) + cos(x)sin(h))/h = sin(x) limh→0 ((cos(h) - 1)/h) + cos(x) limh→0 (sin(h)/h).
From basic trigonometry it is known that
limh→0 ((cos(h) - 1)/h) = 0  and  limh→0 (sin(h)/h) = 1, so that
limh→0 (sin(x)(cos(h) - 1) + cos(x)sin(h))/h = cos(x).


Thank you, Kenneth. This is the trigonometric proof that's needed in this problem.
Your approach to the problem is also very good too!
lim    (sin(x+h)-sin(x))/h

since sin(x+h) = sin(x) cos(h) + cos(x) sin(h) 
then we have
lim   (sin(x) cos(h) + cos(x) sin(h) - sin(x))/h

grouping 1st and 3rd term, we have
lim    ((cos(h) - 1) sin(x) + cos(x) sin(h))/h
Since h not equal 0, a limit exist for the above function.
When h→0  cos(h)=1  and sin(h)=h, the above function becomes
lim   ((1-1)sin(x) + cos(x)(h))/h 
lim   (0 + cos(x)h)/h
lim cos(x)h/h
equation used in this exercise:
sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
lim     cos(x)  = 1       ;     lim   sin(x)  =  x
x→0                                 x→0
applying equations and recognizing pattern of problem is key to solving this kind of math questions, which has been known for some hundreds of year!!!
I bet there is other ways to solve it!!


limx→0sin(x) = 0, not x. However, limx→0(sin(x)/x) = 1.
sin(x) = x - x^3/3! + x^5/5! − x^7/7! + . . .
As x → 0 don't the cube and higher degree terms go to zero much faster than x?
So why can't you say as x → 0, sin(x) → x?
Don't we use the approximation sin(x) ≈ x near x = 0?
The limit of a function f(x) is defined to be a number, not another function. We can't use an approximation in a proof.
(sin(x) - x) = - x^3/3! + x^5/5! − x^7/7! + . . .
As x → 0 the right side → 0, so the value of sin(x) - x → 0 and
sin(x) → x. It's asymptotic.
No, all this means is 
limx→0sin(x) = limx→0x = 0.
Please review the ε-δ-definition of limit.
Also, using the Taylor series of sin(x) in this problem is circular, because to get to the Taylor series you need to take the derivative of sin(x), which is exactly what you are supposed to find in this problem.