Find the limit?

Question

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

 

Answer: cos x

 

Please show all your work.

Answer 1

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).

Answer 2

lim    (sin(x+h)-sin(x))/h

h→0

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

 

 

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

 

cos(x)

 

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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!!