_{h→0}(sin(x)(cos(h) - 1) + cos(x)sin(h))/h = sin(x) lim

_{h→0}((cos(h) - 1)/h) + cos(x) lim

_{h→0}(sin(h)/h).

_{h→0}((cos(h) - 1)/h) = 0 and lim

_{h→0}(sin(h)/h) = 1, so that

_{h→0}(sin(x)(cos(h) - 1) + cos(x)sin(h))/h = cos(x).

Find the limit of (sin(x+h)-sin(x))/h as h approaches 0.

Answer: cos x

Please show all your work.

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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

lim_{h→0} (sin(x)(cos(h) - 1) + cos(x)sin(h))/h = sin(x) lim_{h→0} ((cos(h) - 1)/h) + cos(x) lim_{h→0} (sin(h)/h).

From basic trigonometry it is known that

lim_{h→0} ((cos(h) - 1)/h) = 0 and lim_{h→0} (sin(h)/h) = 1, so that

lim_{h→0} (sin(x)(cos(h) - 1) + cos(x)sin(h))/h = cos(x).

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

h→0

since sin(x+h) = sin(x) cos(h) + cos(x) sin(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

h→0

grouping 1st and 3rd term, we have

h→0

grouping 1st and 3rd term, we have

lim ((cos(h) - 1) sin(x) + cos(x) sin(h))/h

h→0

h→0

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

h→0

lim (0 + cos(x)h)/h

h→0

lim cos(x)h/h

h→0

h→0

cos(x)

---------------------

equation used in this exercise:

sin(a+b) = sin(a)cos(b) + cos(a)sin(b)

and

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!!

lim_{x→0}sin(x) = 0, not x. However, lim_{x→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?

Steve,

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

lim_{x→0}sin(x) = lim_{x→0}x = 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.

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