3 Ways To Evaluate The Limit Of Sin(X)/X As X Approaches 0: L’Hopital’S Rule, Taylor Series, And Squeeze Theorem

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

1

To evaluate the limit lim(sin(x)/x)x -> 0, you can plug x = 0 into the expression sin(x)/x, which results in an indeterminate form of 0/0. However, there are several methods to evaluate this limit.

Method 1: L’Hopital’s Rule
One approach to evaluating this limit is to use L’Hopital’s rule, which states that if the limit of two functions is equal to an indeterminate form, then the limit of the ratio of their derivatives is equal to the original limit. Applying L’Hopital’s rule to the limit in question, we get:

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

Therefore, the value of the limit is 1.

Method 2: Taylor Series Expansion
Another method to evaluate this limit is to use the Taylor series expansion of sin(x) and x as x approaches 0. The Taylor series for sin(x) is:

sin(x) = x – x^3/3! + x^5/5! – …

Similarly, the Taylor series for x is:

x = x

Dividing the Taylor series for sin(x) by the Taylor series for x, we obtain:

sin(x)/x = (x – x^3/3! + x^5/5! – …)/x = 1 – x^2/3! + x^4/5! – …

Taking the limit as x approaches 0, we get:

lim(sin(x)/x)x->0 = lim(1 – x^2/3! + x^4/5! – …)x->0 = 1

Therefore, the value of the limit is 1.

Method 3: Squeeze theorem
Another approach to evaluating this limit is to use the squeeze theorem, which states that if two functions sandwich a third function, and the two surrounding functions have the same limit, then the middle function must also have the same limit. For sin(x)/x, we can observe that:

-1 ≤ sin(x)/x ≤ 1

This is because the maximum value of sin(x) is 1, while the minimum value of sin(x) is -1. Divide each of these values by x, and we get the range of -1/x ≤ sin(x)/x ≤ 1/x. As x approaches 0, the range of -∞ ≤ sin(x)/x ≤ ∞ becomes narrower and narrower, and so the limit of sin(x)/x must be between -1 and 1, inclusive. Since the limit of the cosine function as x approaches 0 is 1, we can set up the following inequality:

cos(x) ≤ sin(x)/x ≤ 1

Taking the limit as x approaches 0, we get:

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

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

In conclusion, the limit lim(sin(x)/x)x->0 is equal to 1, using any of the three methods discussed above.

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