Given that 1 - x2/6 smaller of same as x sin x / 2(1-cos x) smaller or same as 1. Use the Sandwich Theorem to obtain the limit of x sin x / 1 - cos x when x is 0
To find the limit of x sin x / (1 - cos x) as x approaches 0 using the Sandwich Theorem, also known as the Squeeze Theorem, we need to establish two other functions that "sandwich" the given function and have the same limit.
Let's start by simplifying the given function:
x sin x / (1 - cos x)
To simplify further, we can use the trigonometric identity sin(x) = 2sin(x/2)cos(x/2):
x * 2sin(x/2)cos(x/2) / (1 - cos x)
Now, let's consider the denominator (1 - cos x). Notice that as x approaches 0, cos x approaches 1. Thus, the denominator becomes 1 - 1 = 0, which signifies an undefined value.
To resolve this issue, we can multiply the function by (1 + cos x) / (1 + cos x):
(x * 2sin(x/2)cos(x/2) / (1 - cos x)) * ((1 + cos x) / (1 + cos x))
Now, we have:
x * 2sin(x/2)cos(x/2)(1 + cos x) / ((1 - cos x)(1 + cos x))
Simplifying further:
2xsin(x/2)cos(x/2)(1 + cos x) / (1 - cos^2x)
Using the identity cos^2 x = 1 - sin^2 x:
2xsin(x/2)cos(x/2)(1 + cos x) / (1 - (1 - sin^2 x))
Simplifying again:
2xsin(x/2)cos(x/2)(1 + cos x) / sin^2 x
Next, we introduce two other functions, g(x) and h(x), that sandwich the given function and have the same limit as x approaches 0.
Let's define g(x) = 2x(1 + cos x) / sin x and h(x) = 2x(1 - cos x) / sin x.
To verify that g(x) ≤ x sin x / (1 - cos x) ≤ h(x) for all x near 0, we can compare their values.
First, let's consider g(x):
g(x) = 2x(1 + cos x) / sin x
As x approaches 0, cos x approaches 1 and sin x approaches 0.
So, g(x) = 2x(1 + 1) / 0, which is undefined.
Now, let's consider h(x):
h(x) = 2x(1 - cos x) / sin x
As x approaches 0, cos x approaches 1 and sin x approaches 0.
So, h(x) = 2x(1 - 1) / 0, which is also undefined.
Although g(x) and h(x) are undefined at x = 0, we can apply L'Hopital's Rule to find their limits.
Taking the derivatives of g(x) and h(x) with respect to x, we have:
g'(x) = 2(1 + cos x) / cos x
h'(x) = 2(1 - cos x) / cos x
As x approaches 0, the denominator (cos x) approaches 1, so g'(x) and h'(x) become:
g'(0) = 2(1 + 1) / 1 = 4
h'(0) = 2(1 - 1) / 1 = 0
Therefore, the limit as x approaches 0 of both g(x) and h(x) is:
Lim (x -> 0) g(x) = Lim (x -> 0) h(x) = 0
Since the limits of g(x) and h(x) are both 0, we have:
0 ≤ x sin x / (1 - cos x) ≤ 0
Based on the Sandwich Theorem, if f(x) ≤ g(x) ≤ h(x) for all x near 0, and Lim (x -> 0) g(x) = Lim (x -> 0) h(x) = L, then Lim (x -> 0) f(x) = L.
In this case, both g(x) and h(x) have a limit of 0 as x approaches 0.
Therefore, using the Sandwich Theorem, we can conclude that the limit of x sin x / (1 - cos x) as x approaches 0 is also 0.