vanhees71 said:Hint
[tex]x \sin(1/x)=\frac{\sin(1/x)}{1/x}.[/tex]
Mark44 said:I wouldn't. This limit is related to this well-known limit
$$\lim_{t \to 0}\frac{sin(t)}{t}$$
A limit when there is a sine function is the value that a function approaches as the input approaches a specific value. In other words, it is the value that the function "approaches" as we get closer and closer to a certain point on the graph.
To find the limit of a sine function, you can use the following steps:
1. Substitute the value that the input is approaching into the function.
2. Simplify the resulting expression.
3. If the resulting expression is undefined, use algebraic techniques to simplify it.
4. If the resulting expression still cannot be evaluated, the limit does not exist. Otherwise, the value of the expression is the limit.
A one-sided limit only considers the values of the function as the input approaches from one direction (either from the left or right) of the specific value. A two-sided limit, on the other hand, considers the values of the function as the input approaches from both directions of the specific value.
No, a sine function cannot have a limit at a discontinuity. A discontinuity occurs when the function has a break or a hole in its graph, and in this case, the function does not approach a specific value as the input approaches the discontinuity.
Limits of sine functions relate to the unit circle because the values of sine are the y-coordinates of points on the unit circle. The limit of a sine function at a specific value can be determined by looking at the y-coordinate of the point on the unit circle that corresponds to the input value.