What is the limit of (sin x)/x as x approaches 0 and how to prove it?
I know the answer is 1 but my teacher asked me to prove it rigorously. I tried L'Hopital but my teacher said to prove it without L'Hopital since the rule itself depends on this limit. How do I prove it using the squeeze theorem?
1 Answer
KSKaran Singh
✓ Accepted
· 15d ago
▲ 4
Use the geometric squeeze. For 0 < x < pi/2, you can show that sin(x) < x < tan(x) by comparing areas of triangle, sector, and right triangle in a unit circle. Dividing through by sin(x): 1 < x/sin(x) < 1/cos(x). Taking reciprocals reverses inequalities: cos(x) < sin(x)/x < 1. As x approaches 0, cos(x) approaches 1, so by squeeze theorem sin(x)/x approaches 1.
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