- #1
Yankel
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Hello all,
I am looking for the limit of:
sin(x-1) / (x-1)
where x approaches 1. How do I do that ? Thanks !
I am looking for the limit of:
sin(x-1) / (x-1)
where x approaches 1. How do I do that ? Thanks !
Yankel said:Hello all,
I am looking for the limit of:
sin(x-1) / (x-1)
where x approaches 1. How do I do that ? Thanks !
evinda said:We set $x-1=y$.
When $x \to 1$, $y \to 0$.
So, we have the limit:
$$\lim_{y \to 0} \frac{\sin y}{y}$$
and it is known that it is equal to $1$.
You can easily verify it, using L'Hôpital's rule...
The limit of a sine function as x approaches infinity is undefined. This is because the values of sine oscillate between -1 and 1 as x increases without bound, and there is no single value that the function approaches.
To find the limit of a sine function algebraically, you can use the limit definition of sine, which states that the limit of sin(x) as x approaches a is equal to sin(a). This means that the limit of a sine function at a specific point can be found by evaluating the function at that point.
Yes, the limit of a sine function can be negative. This can occur when the function is approaching a point where the sine values are negative, such as at -pi/2, -3pi/2, etc.
The limit of a sine function as x approaches 0 is 0. This can be seen graphically as the function approaches the origin on the x-axis, and the values of sine approach 0.
No, the limit of a sine function may not always be the same as the value of the function at that point. This is because the limit of a function looks at the behavior of the function as it approaches a specific point, while the value of the function at that point is simply the output of the function at that specific input.