Finding the Limit of sin(x) / x as x Approaches 0: L’Hôpital’s Rule vs. The Squeeze Theorem

limit as x approaches 0: sinx/x

To find the limit as x approaches 0 of sin(x) / x, we can use L’Hôpital’s rule or the Squeeze theorem

To find the limit as x approaches 0 of sin(x) / x, we can use L’Hôpital’s rule or the Squeeze theorem.

Using L’Hôpital’s rule:
L’Hôpital’s rule states that for functions of the form f(x) / g(x), if the limit of f(x) / g(x) as x approaches a is of the form 0 / 0 or infinity / infinity, then the limit of the ratio is the same as the limit of the derivative of f(x) divided by the derivative of g(x), as x approaches a.

In this case, let’s consider the ratio f(x) = sin(x) and g(x) = x. The limit as x approaches 0 of f(x) / g(x) is of the form 0 / 0, so we can apply L’Hôpital’s rule:

lim (x -> 0) (sin(x) / x) = lim (x -> 0) (cos(x) / 1)
= cos(0) / 1
= 1

So, the limit as x approaches 0 of sin(x) / x is 1.

Using the Squeeze theorem:
The Squeeze theorem states that if there are two functions, h(x) and g(x), such that h(x) ≤ f(x) ≤ g(x) for all x close to a (except possibly at a), and the limits of h(x) and g(x) as x approaches a are both equal to L, then the limit of f(x) as x approaches a is also L.

In this case, we can use the fact that -1 ≤ sin(x) ≤ 1 for all real numbers x to apply the Squeeze theorem:

-1 ≤ sin(x)/x ≤ 1

As x approaches 0, both -1 and 1 approach 0, so using the Squeeze theorem:

lim (x -> 0) (-1) ≤ lim (x -> 0) (sin(x) / x) ≤ lim (x -> 0) (1)
-1 ≤ lim (x -> 0) (sin(x) / x) ≤ 1

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

Both methods yield the same result, which is 1.

More Answers: