Question

lim cos(pi+h)+1 / h
h->0

evaluate the limit

Answer 1

Relate this to the limit definition of the derivative

Answer 2

i dont get how to solve it

Answer 3

sin pi+sin h + 1 / h or cos pi + cos h / h

Answer 4

Think about lim h -> 0 f(x+h)-f(x)/h, the limit definition of the derivative where f(x) is in this case clearly cos(x)

Answer 5

so what would the answer be

Answer 6

Though I much prefer @Xavier 's method, to provide a non-calculus alternative, multiply but the conjugate\[\frac{\cos(\pi+h)}{\cos(\pi+h)}\]make use of the identity\[\sin^2x+\cos^2x=1\]and the fact that\[\lim_{x\to0}\frac{\sin x}x=1\]and you will arive at the same answer

Answer 7

answer 1 was wrong

Answer 8

the answer is not 1

Answer 9

are you taking pre-calculus or calculus?

Answer 10

calculus

Answer 11

then do this @Xavier 's way

Answer 12

lim = 0

Answer 13

the definition of the derivative of a function at a point a is\[f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}\]

Answer 14

rearrange your expression to look more like this form\[\lim_{h\to0}\frac{\cos (\pi+h)-(-1)}{h}\]how can we write the (-1) part in terms of the cosine?