## limit as x approaches 0: sinx/x

### sinx/x is equal to 1

To evaluate the limit as x approaches 0 of sinx/x, we can use an important trigonometric property. The property states that the limit as x approaches 0 of sinx/x is equal to 1.

However, let’s go through a step-by-step process to understand why this is true.

Step 1: Recall the definition of sinx.

The sine function can be defined as the ratio of the length of the side opposite an angle in a right triangle to the hypotenuse. In this case, we can visualize a right triangle with an angle of x. The opposite side of the triangle would have a length sinx, and the hypotenuse would be 1 (since we are approaching 0, the triangle will become very small).

So sinx/x represents the ratio of the length of the side opposite the angle x to the length of the hypotenuse. This is essentially finding the ratio of the length of the opposite side to the hypotenuse.

Step 2: Use the Squeeze Theorem.

The Squeeze Theorem states that if we have three functions, f(x), g(x), and h(x), where for all values of x close to a certain point, f(x) ≤ g(x) ≤ h(x), and the limits as x approaches a of f(x) and h(x) are both L, then the limit as x approaches a of g(x) is also L.

In our case, we can rewrite sinx/x as (sinx)/x and evaluate the limit as x approaches 0.

Step 3: Applying the Squeeze Theorem.

We know that for all x ≠ 0, -1 ≤ sinx ≤ 1. So we have:

-1/x ≤ sinx/x ≤ 1/x.

Take the limit as x approaches 0 for all three sides of the inequality. We already know the limits of the two outermost expressions:

lim (x → 0) (-1/x) = -∞ and lim (x → 0) (1/x) = ∞.

Since -∞ ≤ sinx/x ≤ ∞, and the limits of the outer expressions are both infinite, we can apply the Squeeze Theorem to conclude that the limit as x approaches 0 of sinx/x must also be infinite.

However, as mentioned earlier, we know that the limit as x approaches 0 of sinx/x is equal to 1. This result is established using more advanced concepts, such as L’Hôpital’s Rule or Taylor series expansion. These methods involve derivatives and power series, but they are beyond the scope of this explanation.

In summary, the limit as x approaches 0 of sinx/x is equal to 1.

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