How To Evaluate The Limit Of Sin(X)/X Using The Squeeze Theorem

lim x->0 sinx/x

1

To evaluate this limit, we can use the squeeze theorem.

First, we know that sin(x) is between -1 and 1 for all values of x. Therefore:

-1 ≤ sin(x) ≤ 1

Next, we can divide both sides by x:

-1/x ≤ sin(x)/x ≤ 1/x

As x approaches 0, 1/x becomes very large in magnitude. Therefore, the bound on the right side of the inequality becomes very loose, and we can ignore it.

We’re left with:

-1/x ≤ sin(x)/x

Now, we can take the limit as x approaches 0:

lim x→0 (-1/x) ≤ lim x→0 (sin(x)/x)

As x approaches 0, -1/x approaches negative infinity.

To evaluate the right side of the inequality, we can use L’Hopital’s rule:

lim x→0 (sin(x)/x) = lim x→0 (cos(x)) = cos(0) = 1

Therefore, we have:

negative infinity ≤ lim x→0 (sin(x)/x) ≤ 1

Since the limit is sandwiched between two values that approach negative and positive infinity, respectively, we can conclude that the limit does not exist.

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