Understanding the Limit of Sin(x)/x as x Approaches 0: A Simplified Explanation and Proof

limit as x approaches 0: sinx/x

To evaluate the limit of sin(x)/x as x approaches 0, we can’t simply substitute the value 0 into the expression because it results in an indeterminate form of 0/0

To evaluate the limit of sin(x)/x as x approaches 0, we can’t simply substitute the value 0 into the expression because it results in an indeterminate form of 0/0. Instead, we need to find a way to simplify the expression to be able to evaluate the limit.

We can use the property of the sine function that states lim(x->0) sin(x)/x = 1. This property can be proven by using the concept of the squeeze theorem, but it is commonly known and widely used.

To prove this property, we can use the fact that the sine function is bounded by -1 and 1, so we have -1 ≤ sin(x) ≤ 1 for all values of x. Dividing the inequality by x (assuming x is positive), we get -1/x ≤ sin(x)/x ≤ 1/x. Now, as x approaches 0, both -1/x and 1/x approach ± infinity, but sin(x)/x remains bounded between these two values. Therefore, by the squeeze theorem, the limit of sin(x)/x as x approaches 0 is equal to 1.

Therefore, the solution to the limit as x approaches 0 of sin(x)/x is 1.

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