Question

does someone knows how to do this limit?

Answer 1

let a = x+h

Answer 2

you seem to be trying to find the derivative of cos(x), or something similar

Answer 3

no....the exercise is just get the lmit....i will try with the substitution that you make

Answer 4

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

which them allows us to split the fraction

Answer 5

\[\frac{cos(x)-cos(x)cos(h)+sin(x)sin(h)}{x-(x+h)}\]
\[\frac{cos(x)-cos(x)cos(h)+sin(x)sin(h)}{-h}\]
\[\frac{cos(x)cos(h)-cos(x)-sin(x)sin(h)}{h}\]
\[\frac{cos(x)[cos(h)-1]-sin(x)sin(h)}{h}\]
\[\lim_{h\to 0}~~cos(x)\frac{[cos(h)-1]}{h}-sin(x)\frac{sin(h)}{h}\]

Answer 6

ok.....i need a lot of time to understand this....thanks @amistre64

Answer 7

i can do it from here...you did the hard part

Answer 8

its just an idea, might work but we would need to know beforehand that the cos h parts go to 0 and the sin h parts got to 1