Which limit equals 0 because the numerator is bounded while the denominator grows without bound: lim_{x→∞} sin x / x?

Which limit equals 0 because the numerator is bounded while the denominator grows without bound: lim_{x→∞} sin x / x?

• A. 0
• B. 1
• C. ∞
• D. -1

Answer: (A)

When the numerator stays bounded while the denominator grows without bound, the fraction must go to zero. The sine function is always between -1 and 1, so for all x we have |sin x| ≤ 1. This gives |sin x / x| ≤ 1/|x|. As x approaches infinity, 1/|x| tends to 0, so the whole expression sin x / x must approach 0 as well. The oscillations of sin x fade away because they’re being divided by an unboundedly large number. Hence the limit is 0.

When the numerator stays bounded while the denominator grows without bound, the fraction must go to zero. The sine function is always between -1 and 1, so for all x we have |sin x| ≤ 1. This gives |sin x / x| ≤ 1/|x|. As x approaches infinity, 1/|x| tends to 0, so the whole expression sin x / x must approach 0 as well. The oscillations of sin x fade away because they’re being divided by an unboundedly large number. Hence the limit is 0.