Exploring the Limit of Sin(x)/x as x Approaches 0: Insights from Trigonometric Identity and Squeeze Theorem

lim x->0 sinx/x

To evaluate the limit as x approaches 0 for sin(x)/x, we can use a basic trigonometric identity and the concept of the squeeze theorem

To evaluate the limit as x approaches 0 for sin(x)/x, we can use a basic trigonometric identity and the concept of the squeeze theorem.

The trigonometric identity we’ll use is:
lim θ->0 (sin θ)/θ = 1

Using this identity, we see that sin(x)/x is equivalent to sin(x)/x * (1/sin(x)) = 1/x.

Now, let’s examine the behavior of 1/x as x approaches 0. We can use the squeeze theorem to do this. The squeeze theorem states that if we have three functions, f(x), g(x), and h(x), such that f(x) ≤ g(x) ≤ h(x) for all x in some interval (except possibly x = a), and lim x->a f(x) = L = lim x->a h(x), then lim x->a g(x) = L as well.

In this case, we can choose f(x) = -1/x and h(x) = 1/x. Notice that for all x ≠ 0, we have -1/x ≤ 1/x ≤ 1/x. As x approaches 0, both -1/x and 1/x approach negative and positive infinity, respectively.

Therefore, by the squeeze theorem, since -1/x ≤ 1/x ≤ 1/x and lim x->0 -1/x = -∞ and lim x->0 1/x = +∞, we can conclude that lim x->0 1/x is equal to +∞.

Hence, the limit of sin(x)/x as x approaches 0 is infinity, or formally written as:
lim x->0 sin(x)/x = +∞.

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