Question

If f(x) = sin(x), then lim(x→2π) [f(2π) - f(x)] / [x - 2π] equals:

a) 0
b) 1
c) -1
d) [infinity]

Answer 1

The limit
`lim(x→2π) [f(2π) - f(x)] / [x - 2π]` evaluates as
`-1` using L'Hôpital's Rule.

Step-by-step explanation:

The limit in question,
`lim(x→2π) [f(2π) - f(x)] / [x - 2π]` , evaluates how the function
`f(x) = sin(x)` changes as x approaches
`2π`. Notably, due to the periodic nature of the sine function, we know that
`sin(2π) = 0`.

Therefore, the expression simplifies to
`lim(x→2π) [-sin(x)] / [x - 2π]`.

By using L'Hôpital's Rule, we can differentiate the numerator and the denominator separately:

• The derivative of the numerator
  `-sin(x)` with respect to
  `x is -cos(x)` .
• The derivative of the denominator
  `x - 2π` with respect to
  `x is 1` .

After differentiation, we have
`lim(x→2π) -cos(x)` which is simply
`-cos(2π)` . Since
`cos(2π) = 1` the limit evaluates to
`-1`.