Finding the Limit of sin(x)/x as x Approaches 0: Concept, Trigonometric Identities, and Squeeze Theorem

lim x->0 sinx/x

To find the limit of sin(x)/x as x approaches 0, we can use the concept of limits and apply some trigonometric identities

To find the limit of sin(x)/x as x approaches 0, we can use the concept of limits and apply some trigonometric identities.

First, let’s recall the identity sin(x)/x = 1 when x is exactly 0. However, when we let x approach 0, we encounter the issue of dividing by 0, which is undefined. Therefore, we need to use a different approach.

We can rewrite sin(x)/x as (sin(x))/(x) * (x)/1. The term (x)/1 is equal to x and can be plotted on a graph, forming a line with a slope of 1 passing through the origin. As x approaches 0, this line gets closer and closer to the x-axis.

Next, let’s consider the term (sin(x))/(x). We know that sin(0) = 0, so substituting 0 into the term would result in (0)/(0), which is again undefined. However, we can use a trigonometric identity to simplify this term.

By the Squeeze Theorem, we can state that -1 ≤ sin(x) ≤ 1 for all x. Dividing every term in this inequality by x, we get -1/x ≤ sin(x)/x ≤ 1/x.

Now, as x approaches 0, both -1/x and 1/x approach ∞ or -∞, depending on the sign of x. This means that sin(x)/x is “squeezed” between -∞ and ∞.

Therefore, by the Squeeze Theorem, the limit of sin(x)/x as x approaches 0 is equal to 1.

In mathematical notation:
lim x->0 sin(x)/x = 1

So, when x gets infinitely close to 0, the value of sin(x)/x gets closer and closer to 1.

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