How to Prove the Fundamental Limit in Calculus: The Squeeze Theorem for Evaluating the Limit of (sinx/x) when x→0.

limit as x approaches 0: sinx/x

1

To evaluate this limit, we can use the following trigonometric identity:

lim x→0 (sinx/x) = 1

This identity is a fundamental limit in calculus and is often referred to as the Squeeze Theorem.

To prove this identity, we can use the following steps:

Step 1: Use the trigonometric identity sinx = x – (x^3/3!) + (x^5/5!) – … + (-1)^n (x^(2n+1))/(2n+1)! + …

Step 2: Substituting sinx = x – (x^3/3!) + (x^5/5!) – … + (-1)^n (x^(2n+1))/(2n+1)! + …, we can rewrite the original limit as:

lim x→0 (sinx/x) = lim x→0 1 – (x^2/3!) + (x^4/5!) – … + (-1)^n (x^(2n))/(2n+1)! + …

Step 3: Recall that the series expansion for e^x = 1 + x + (x^2/2!) + (x^3/3!) + …

Step 4: Substituting x^2 for x in the series expansion for e^x, we obtain:

e^(x^2) = 1 + x^2 + (x^4/2!) + (x^6/3!) + …

Step 5: Multiplying the series expansion for e^(x^2) by (sinx)/x, we obtain:

(e^(x^2)) (sinx/x) = (1 + x^2/1!) + (x^3/3!)(1 + x^2/2!) + (x^4/4!)(1 + x^2/3!) + …

Step 6: Since all the terms in the series expansion for e^(x^2) are positive for x ≠ 0, we can see that:

(e^(x^2)) (sinx/x) > 1

Step 7: Dividing both sides of the inequality by e^(x^2), we obtain:

(sin x/x) > e^(-x^2)

Step 8: Since e^(-x^2) approaches 1 as x approaches 0, we can see that:

1 > lim x→0 (sinx/x)

Step 9: Since the limit of (sinx/x) exists and equals 1-by steps 8 and identity earlier stated in step 1, we can conclude that:

lim x→0 (sinx/x) = 1

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