Use the Sandwich Theorem to obtain the limit of 2 Sin x / x when x is 0

To evaluate the limit of 2sin(x)/x as x approaches 0 using the Sandwich Theorem (also known as the Squeeze Theorem or Pinching Theorem), we need to find two functions that "sandwich" the given function and whose limits are equal as x approaches 0.

The Sandwich Theorem states that if f(x) ≤ g(x) ≤ h(x) for all x in a certain interval (except possibly at x = c), and the limits of f(x) and h(x) as x approaches c are both L, then the limit of g(x) as x approaches c is also L.

In this case, we have the function f(x) = -2/x and h(x) = 2/x. Let's find their limits as x approaches 0:

1. Limit of f(x) as x approaches 0:
lim[x->0] (-2/x) = negative infinity (∞)

2. Limit of h(x) as x approaches 0:
lim[x->0] (2/x) = positive infinity (∞)

Since f(x) approaches negative infinity and h(x) approaches positive infinity as x approaches 0, we need to find a function g(x) that lies between f(x) and h(x) and has the same limit as x approaches 0.

Now, let's consider the function g(x) = 2sin(x)/x. To apply the Sandwich Theorem, we need to prove that g(x) is always between f(x) and h(x) for x near 0.

Since -1 ≤ sin(x) ≤ 1 for all x, we can rewrite g(x) as:
-2/x ≤ 2sin(x)/x ≤ 2/x

Now, let's find the limits of f(x), g(x), and h(x) as x approaches 0:

1. Limit of f(x) as x approaches 0:
lim[x->0] (-2/x) = negative infinity (∞)

2. Limit of g(x) as x approaches 0:
There is a well-known limit:
lim[x->0] (sin(x)/x) = 1

We multiply this limit by 2:
lim[x->0] (2(sin(x)/x)) = 2(1) = 2

3. Limit of h(x) as x approaches 0:
lim[x->0] (2/x) = positive infinity (∞)

Since f(x) approaches negative infinity and h(x) approaches positive infinity, and g(x) is always between f(x) and h(x), we can conclude that the limit of g(x) as x approaches 0 is also 2.

Therefore, the limit of 2sin(x)/x as x approaches 0 is 2.