# How do you use the Squeeze Theorem to find #lim Sin(x)/x# as x approaches zero?

For a non-rigorous proof, please see below.

For a positive central angle of

Source:

The geometric idea is that

So we have:

For small positive

So

so

We also have, for these small

so

Since both one sided limits are

Note

This proof uses the fact that

That fact can be proved from the fact that

Which can be proved using the squeeze theorem in a argument rather like the one used above.

Furthermore: Using both of those facts we can show that the sine and cosine functions are continuous at every real number.

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To use the Squeeze Theorem to find the limit of Sin(x)/x as x approaches zero, we can establish two other functions that "squeeze" the original function.

First, we know that -1 ≤ Sin(x) ≤ 1 for all x. Therefore, we can say that -1/x ≤ Sin(x)/x ≤ 1/x for all x ≠ 0.

Next, we can take the limit as x approaches zero for each of these inequalities.

As x approaches zero, -1/x approaches negative infinity, and 1/x approaches positive infinity.

Since -1/x ≤ Sin(x)/x ≤ 1/x for all x ≠ 0, and the limits of -1/x and 1/x as x approaches zero are both infinity, we can conclude that the limit of Sin(x)/x as x approaches zero is also infinity.

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