However, along the path y = - sin ⁡ x, which lies in the domain of f ⁢ ( x, y ) for all x ≠ 0, the limit does not exist. Along any line y = m ⁢ x in the domain of the f ⁢ ( x, y ), the limit is 0. Step back and consider what we have just discovered. = “ - 2 / 0 ” ⇒ the limit does not exist. Mathematically, a function f(x) is continuous at a point x a. A continuous function is a function which when drawn on a paper does not have a break. = lim x → 0 ⁡ - sin ⁡ ( - x ⁢ sin ⁡ x ) ⁢ ( - sin ⁡ x - x ⁢ cos ⁡ x ) 2 + cos ⁡ ( - x ⁢ sin ⁡ x ) ⁢ ( - 2 ⁢ cos ⁡ x + x ⁢ sin ⁡ x ) sin ⁡ x Continuity of functions is an important unit of Calculus as it forms the base and it helps us further to prove whether a function is differentiable or not. = lim x → 0 ⁡ cos ⁡ ( - x ⁢ sin ⁡ x ) ⁢ ( - sin ⁡ x - x ⁢ cos ⁡ x ) 1 - cos ⁡ x ( “ = 0 / 0 ⁢ ” ) = lim x → 0 ⁡ sin ⁡ ( - x ⁢ sin ⁡ x ) x - sin ⁡ x Lim ( x, - sin ⁡ x ) → ( 0, 0 ) ⁡ sin ⁡ ( - x ⁢ sin ⁡ x ) x - sin ⁡ x Now consider the limit along the path y = - sin ⁡ x: This line is not in the domain of f, so we have found the following fact: along every line y = m ⁢ x in the domain of f, lim ( x, y ) → ( 0, 0 ) ⁡ f ⁢ ( x, y ) = 0. īy applying L’Hôpital’s Rule, we can show this limit is 0 except when m = - 1, that is, along the line y = - x. Conditions (a) and (b) are technically contained implicitly in (c), but we state them explicitly to emphasize their individual importance. (a) f has a limit as x a, (b) f is defined at x a, and. A function f is continuous at x a provided that. = lim x → 0 ⁡ sin ⁡ ( m ⁢ x 2 ) x ⋅ 1 m + 1. More formally, we make the following definition. = lim x → 0 ⁡ sin ⁡ ( m ⁢ x 2 ) x ⁢ ( m + 1 ) First, however, consider the limits found along the lines y = m ⁢ x as done above. We are to show that lim ( x, y ) → ( 0, 0 ) ⁡ f ⁢ ( x, y ) does not exist by finding the limit along the path y = - sin ⁡ x. Let f ⁢ ( x, y ) = sin ⁡ ( x ⁢ y ) x + y.

Define continuity calculus