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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