Use pinching theorem to evaluate lim x-->1 ((x-1)sin(1/x-1))
I'm confused in the pinch theorem analytically
note that if u = 1/(x-1) then what you have is
sin(u)/u
You have probably seen that this limit is 1, so follow the same argument.
Or, try google. A good discussion is at
math.ucsb.edu/~jcs/SqueezeTheorem.pdf
Would it be 0?
you are correct. 1/(x-1) -> ∞
I tried to pound a round peg into a square hole.
To evaluate the limit in question using the Pinching Theorem (also known as the Squeeze Theorem), we need to find two functions that "squeeze" the expression ((x-1)sin(1/x-1)) as x approaches 1.
First, let's analyze the expression ((x-1)sin(1/x-1)) as x approaches 1. Notice that the expression is undefined at x = 1 since we would be dividing by zero in the denominator. However, we can still consider the limit as x approaches 1.
To apply the Pinching Theorem analytically, we need to find two functions g(x) and h(x) that satisfy the following conditions:
1. g(x) ≤ ((x-1)sin(1/x-1)) ≤ h(x) for all x in some open interval around 1, except possibly at x = 1.
2. lim[x->1] g(x) = lim[x->1] h(x) = L (some number)
Once we find such functions, we can conclude that lim[x->1] ((x-1)sin(1/x-1)) = L.
Now, let's determine the functions g(x) and h(x). Since we are interested in the behavior of ((x-1)sin(1/x-1)) as x approaches 1, we can simplify the expression by substituting y = 1/x-1. This leads to:
((x-1)sin(1/x-1)) = (1/y)sin(y)
With this substitution, we can rewrite the expression as:
((x-1)sin(1/x-1)) = (sin(y))/y
Now, let's consider two functions: g(x) = 1 and h(x) = sin(y)/y.
1. g(x) = 1 is a constant function that is always less than or equal to ((x-1)sin(1/x-1)).
2. To prove that h(x) = sin(y)/y satisfies the second condition, we need to find its limit as x approaches 1. By substituting y = 1/x-1, the limit of h(x) as x approaches 1 becomes:
lim[x->1] h(x) = lim[x->1] (sin(y))/y
To evaluate this limit, we can use the fact that sin(y)/y approaches 1 as y approaches 0. Therefore:
lim[x->1] h(x) = lim[y->0] (sin(y))/y = 1
Since both g(x) = 1 and h(x) = sin(y)/y satisfy the conditions of the Pinching Theorem, we can conclude that:
lim[x->1] ((x-1)sin(1/x-1)) = lim[x->1] (sin(y))/y = 1
Therefore, the limit of ((x-1)sin(1/x-1)) as x approaches 1 is equal to 1.